Statistics Quizzes

Let's break down the calculation and explanation for \(P(X \leq 3)\) when \(X\) is a binomial random variable with parameters \(n = 10\) (number of trials) and \(p = 0.3\) (probability of success on each trial).

### Binomial Distribution

The binomial distribution describes the probability of obtaining exactly \(k\) successes out of \(n\) trials, with each trial having two possible outcomes: success (with probability \(p\)) or failure (with probability \(1-p\)).

The probability mass function (PMF) for a binomial distribution is given by:

\[ P(X = k) = \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} \]

### Calculating \(P(X \leq 3)\)

To find \(P(X \leq 3)\), we sum the probabilities of \(X\) being 0, 1, 2, and 3:

\[ P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) \]

Let's calculate each component:

1. **\(P(X=0)\):**

\[ P(X = 0) = \frac{10!}{0!(10-0)!} (0.3)^0 (0.7)^{10} = (0.7)^{10} \]

2. **\(P(X=1)\):**

\[ P(X = 1) = \frac{10!}{1!(10-1)!} (0.3)^1 (0.7)^9 = 10 \times 0.3 \times (0.7)^9 \]

3. **\(P(X=2)\):**

\[ P(X = 2) = \frac{10!}{2!(10-2)!} (0.3)^2 (0.7)^8 = 45 \times (0.3)^2 \times (0.7)^8 \]

4. **\(P(X=3)\):**

\[ P(X = 3) = \frac{10!}{3!(10-3)!} (0.3)^3 (0.7)^7 = 120 \times (0.3)^3 \times (0.7)^7 \]

Summing up these probabilities gives us \(P(X \leq 3)\).

### Conclusion

Using the factorial notation to express binomial coefficients, we've analytically shown how to calculate the cumulative probability \(P(X \leq 3)\) for a binomial random variable \(X\) with parameters \(n = 10\) and \(p = 0.3\), which we previously found to be approximately 0.650. This method sums the probabilities of getting up to 3 successes in 10 trials, demonstrating the application of binomial distribution principles.

Given the following data:
- Mean lifespan of light bulbs, \(\mu = 1200\) hours
- Standard deviation, \(\sigma = 100\) hours
- Sample size, \(n = 40\)
- Sample mean to compare against, \(\bar{x} = 1210\) hours

We aim to find the probability that the average lifespan of bulbs in the sample will be more than 1210 hours.

First, we calculate the z-score using the formula:
\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]

Plugging in the given values:
\[ z = \frac{1210 - 1200}{\frac{100}{\sqrt{40}}} \approx 0.63 \]

Next, to find the probability that the average lifespan is more than 1210 hours, we look for the area to the right of this z-score in the standard normal distribution. This probability, \(P\), is given by:
\[ P = 1 - \Phi(z) \]

Where \(\Phi(z)\) is the cumulative distribution function (CDF) of the standard normal distribution evaluated at z.

Given our calculated z-score:
\[ P = 1 - \Phi(0.63) \approx 0.264 \]

Therefore, the probability that the average lifespan of the bulbs in the sample will be more than 1210 hours is approximately 0.264.

The correct answer is B) 0.2635.

A random variable is a numerical quantity that results from a random experiment, where the outcome is uncertain. It assigns a numerical value to each outcome in the sample space of a stochastic (or random) process. The key characteristic of a random variable is its ability to take different values based on the inherent randomness of the process being observed. Unlike observable, numerical, expected, or experimental variables, a random variable specifically captures the concept of uncertainty and variability in outcomes, making it a fundamental concept in probability and statistics.

For large sample hypothesis tests concerning a single population proportion, the test statistic follows a standard normal distribution, especially as the sample size increases, making the approximation to the standard normal distribution more accurate due to the central limit theorem.

The complement of a set \(B\) within a sample space \(S\) consists of all the elements in \(S\) that are not in \(B\). Given the sample space \(S = \{a, b, c, d, e\}\) and the set \(B = \{b, d\}\), the complement of \(B\) includes all elements in \(S\) except for those in \(B\).

Therefore, the complement of \(B\) is \(\{a, c, e\}\), since these are the elements in the sample space \(S\) that are not in \(B\). This means that option d is correct, illustrating the concept of set complement where you include all elements of the universal set (in this case, the sample space \(S\)) that are not part of the set in question (\(B\)).

The mean (or expected value) of a binomial random variable is given by the formula:
\[ \mu = n \cdot p \]
where:
- \(n\) is the number of trials,
- \(p\) is the probability of success on each trial.

Given the parameters \(n = 10\) and \(p = 0.3\), we can calculate the mean as follows:
\[ \mu = 10 \cdot 0.3 = 3 \]

Therefore, the expected value (mean) of a binomial distribution with these parameters is 3. This means that, on average, we can expect 3 successes out of 10 trials if each trial has a probability of success of 0.3. This aligns with option b, making it the correct answer.

The probability \(P(-2.05 < Z < 1.96)\) for a standard normal variable \(Z\) represents the area under the standard normal distribution curve between the z-scores of \(-2.05\) and \(1.96\). The standard normal distribution is symmetric around zero, and the areas (probabilities) associated with specific z-score ranges can be found using the standard normal distribution table or a calculator.

The z-scores of \(-2.05\) and \(1.96\) roughly correspond to the 2.5th percentile and the 97.5th percentile of the distribution, respectively. The total area (probability) between these two z-scores encompasses most of the distribution, leaving about 2.5% in the tails outside of this range. Therefore, the combined area under the curve between \(-2.05\) and \(1.96\) is approximately 95% of the total area, which corresponds to a probability of 0.9548.

This calculation is based on the properties of the standard normal distribution and is a common way to approximate probabilities for normally distributed data. The exact values may vary slightly based on the precision of the tables or computational tools used, but the option closest to the theoretical value is the correct answer.

The mean of the sample proportion, often denoted as \(\hat{p}\), for random samples is equal to the population proportion \(p\). This is because the expected value (mean) of the sample proportion is the same as the population proportion from which the samples are drawn.

Given that the population proportion \(p = 0.35\), regardless of the sample size (in this case, \(n = 900\)), the mean of the sample proportion \(\hat{p}\) will also be \(0.35\). This principle is rooted in the idea that, on average, the sample will accurately reflect the population characteristics if the sample is random and the size is sufficiently large.

Therefore, the correct answer is \(0.35\), which indicates that the mean of the sample proportion obtained from random samples of size 900 is expected to be equal to the population proportion of \(0.35\).

A statistic is a numerical value that summarizes or describes an aspect of a sample of data. It is derived from the sample data itself and is used to make inferences about the population from which the sample was drawn. Unlike a parameter, which describes a characteristic of the entire population, a statistic specifically refers to the characteristics of the sample. Examples of statistics include the sample mean, sample variance, and sample standard deviation, among others. Therefore, the term "statistic" best fits the description of a number that summarizes some aspect of the sample data.

The standard error (SE) of the sample proportion is calculated using the formula:
\[ SE = \sqrt{\frac{p(1-p)}{n}} \]
where \(p\) is the population proportion, and \(n\) is the sample size.

Given:
- \(p = 0.48\)
- \(n = 169\)

Plugging these values into the formula gives:
\[ SE = \sqrt{\frac{0.48(1-0.48)}{169}} \]

After calculating, the standard error (SE) of the sample proportion is approximately \(0.038\), which when rounded to two decimal places is \(0.04\).

In probability theory, an "event" refers to any subset of a sample space that can occur when a random experiment is performed. It is a collection of possible outcomes from the sample space. For example, in a dice-rolling experiment, the sample space is all possible outcomes \(\{1, 2, 3, 4, 5, 6\}\), and an event could be rolling an even number, represented as the subset \(\{2, 4, 6\}\). Therefore, an event is not just a random selection but a specific outcome or set of outcomes that we are interested in observing within the framework of probability and statistics. This makes option c, "event," the correct terminology for describing a subset of a sample space in the context of probability theory.

The expression \(P(Z \leq z) = 0.7734\) refers to the probability that the standard normal variable \(Z\) is less than or equal to a certain z-value, which in this case corresponds to a probability of 0.7734. This probability is associated with the area under the standard normal distribution curve to the left of the specified z-value.

To find the correct z-value, one typically refers to a z-table (standard normal distribution table) or uses a statistical calculator that can invert this probability into a z-score. The value of 0.7734 suggests a z-score that is to the right of the mean (since it's more than 0.5, indicating more than half of the area under the curve is to the left of this point).

A z-score of 0.75 is approximately where the cumulative probability from the left up to this z-score matches 0.7734 in the standard normal distribution. Therefore, the correct z-value that satisfies \(P(Z \leq z) = 0.7734\) is 0.75, indicating that about 77.34% of the distribution falls to the left of this point on the z-distribution curve. This methodology is based on the properties of the standard normal distribution, where probabilities are mapped to z-scores that represent standard deviations from the mean.

When two distinguishable coins are tossed, each coin has two possible outcomes: heads (h) or tails (t). Because the coins are distinguishable, the outcome of the first coin can be combined with the outcome of the second coin in every possible way, leading to four unique combinations:

- Both coins show heads (hh).
- Both coins show tails (tt).
- The first coin shows heads and the second coin shows tails (ht).
- The first coin shows tails and the second coin shows heads (th).

Thus, the complete sample space that represents all possible outcomes is {hh, tt, ht, th}.

To find the probability that Lara's order contains one or more defective pieces, we can use the concept of the Poisson distribution, given the average rate of defective mugs.

The Poisson distribution is appropriate when events occur randomly in time or space, and the average rate of occurrence is known. In this case, we're interested in the number of defective mugs in Lara's order, given the average rate provided by the online site.

The Poisson distribution probability mass function is given by:

\[ P(X = k) = \frac{{e^{-\lambda} \cdot \lambda^k}}{{k!}} \]

Where:
- \( P(X = k) \) is the probability of observing \( k \) events,
- \( e \) is Euler's number (approximately 2.71828),
- \( \lambda \) is the average rate of occurrence,
- \( k \) is the number of events.

Given:
- Average rate of defective mugs (\( \lambda \)) = 5 mugs per 50,000 produced.

First, we need to adjust the average rate to the number of mugs Lara ordered. Since Lara ordered 5 mugs, the adjusted average rate for her order is \( \lambda = \frac{5}{50000} \times 5 = \frac{1}{10000} \).

Now, we can calculate the probability of Lara receiving one or more defective mugs:

\[ P(X \geq 1) = 1 - P(X = 0) \]

\[ P(X \geq 1) = 1 - \frac{{e^{-\frac{1}{10000}} \cdot \left(\frac{1}{10000}\right)^0}}{{0!}} \]

\[ P(X \geq 1) = 1 - e^{-\frac{1}{10000}} \]

Using a calculator, we find:

\[ P(X \geq 1) \approx 1 - 0.9999000049998333 \]

\[ P(X \geq 1) \approx 0.00009999500016672938 \]

So, the probability that Lara's order contains one or more defective pieces is approximately \( 0.0001 \), or \( 0.01\% \).

When rolling two fair six-sided dice, the total number of outcomes is \(6 \times 6 = 36\), as there are 6 possible outcomes for each die.

Now, let's consider the outcomes where doubles are rolled. There are 6 possible doubles: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6).

So, the probability of rolling doubles is the number of outcomes with doubles divided by the total number of outcomes:

\[ P(\text{doubles}) = \frac{\text{number of outcomes with doubles}}{\text{total number of outcomes}} = \frac{6}{36} = \frac{1}{6} \]

To find the probability of not rolling doubles, we can subtract the probability of rolling doubles from 1, because if an event doesn't happen, its complement happens:

\[ P(\text{not doubles}) = 1 - P(\text{doubles}) = 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6} \]

So, the probability of not rolling doubles is \( \frac{5}{6} \).

To find the percentage of people who score between 52 and 148 on IQ tests using the empirical rule, we first need to determine how many standard deviations away from the mean these scores are.

Given:
- Mean (μ) = 100
- Standard deviation (σ) = 16

For the score of 52:
\[ Z_1 = \frac{52 - 100}{16} = \frac{-48}{16} = -3 \]

For the score of 148:
\[ Z_2 = \frac{148 - 100}{16} = \frac{48}{16} = 3 \]

Now, using the empirical rule:
- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.

Since both 52 and 148 fall within three standard deviations of the mean (one is 3 standard deviations below the mean and the other is 3 standard deviations above), we can expect a large percentage of the population to fall within this range.

Using the empirical rule, we know that approximately 99.7% of the data falls within three standard deviations of the mean. Therefore, the percentage of people who score between 52 and 148 on IQ tests is approximately 99.7%.

The second quartile (Q2) or median of a data set is the value that divides the set into two equal halves. To find Q2, the data set must first be sorted in ascending order. Once sorted, if the number of observations is odd, Q2 is the middle value. If the number of observations is even, Q2 is the average of the two middle values.

After sorting the given data set and calculating the median, the second quartile (Q2) of the data set {1.5, 2.3, 1.7, 3.6, 2.9, 4.2, 3.3} is found to be 2.9.

Therefore, the correct answer is: b. 2.9.

To estimate the minimum sample size needed to compare the means of two populations with a specified confidence level and margin of error, we can use the formula for the sample size in a two-sample confidence interval for the difference in means.

The formula for the sample size needed in each group is given by:

\[ n = \left( \frac{{Z_{\alpha/2} \cdot \sigma_1}}{{E}} \right)^2 + \left( \frac{{Z_{\alpha/2} \cdot \sigma_2}}{{E}} \right)^2 \]

Where:
- \( n \) = sample size in each group
- \( Z_{\alpha/2} \) = critical value for the desired level of confidence (for 98% confidence level, \( Z_{\alpha/2} = 2.33 \))
- \( \sigma_1 \) = standard deviation of the first group
- \( \sigma_2 \) = standard deviation of the second group
- \( E \) = margin of error

Given:
- \( E = 6 \)
- \( Z_{\alpha/2} = 2.33 \)
- \( \sigma_1 = 11.50 \)
- \( \sigma_2 = 14.20 \)

Substituting the values into the formula:

\[ n = \left( \frac{{2.33 \cdot 11.50}}{{6}} \right)^2 + \left( \frac{{2.33 \cdot 14.20}}{{6}} \right)^2 \]

\[ n = \left( \frac{{26.795}}{{6}} \right)^2 + \left( \frac{{33.086}}{{6}} \right)^2 \]

\[ n = (4.465)^2 + (5.514)^2 \]

\[ n = 19.946225 + 30.415796 \]

\[ n \approx 50.361021 \]

Since you can't have a fraction of a sample, you would need to round up to the nearest whole number. Therefore, you would need at least 51 samples from each flavor to estimate the difference in average sales with a 98% confidence level and a margin of error of $6.

To determine the probability of rolling a face showing two spots exactly twice when a die is rolled eight times, we can use the binomial probability formula. The binomial probability formula is given by:

\[ P(X = k) = {n \choose k} \cdot p^k \cdot (1 - p)^{n-k} \]

where:
- \(P(X = k)\) is the probability of getting \(k\) successes (in this case, rolling a two) out of \(n\) trials (die rolls),
- \( {n \choose k} \) is the number of combinations of \(n\) items taken \(k\) at a time,
- \(p\) is the probability of success on any given trial (rolling a two on a fair six-sided die has a probability of \(\frac{1}{6}\)),
- \(n\) is the number of trials (in this case, 8 rolls),
- \(k\) is the number of successes (in this case, rolling a two twice),
- \(1 - p\) is the probability of failure on any given trial.

Let's plug in the numbers:
- \(n = 8\) (eight rolls),
- \(k = 2\) (rolling a two twice),
- \(p = \frac{1}{6}\) (probability of rolling a two).

Now, we calculate the probability.

The probability that, when a die is rolled eight times, the face showing two spots appears exactly twice is approximately 0.2605 or 26.05%.

The deterministic part of the simple linear regression equation captures the trend in the dependent variable (\( Y \)) as the independent variable (\( X \)) increases. This trend is described by the slope coefficient \( \beta_1 \), which indicates the rate of change in \( Y \) for a unit increase in \( X \). So, option d accurately describes the deterministic part of the simple linear regression equation.

To estimate the minimum sample size needed to achieve a specific level of confidence and margin of error for comparing two proportions, we can use the formula for sample size in a two-sample confidence interval for the difference in proportions.

The formula for the sample size needed in each group is given by:

\[ n = \frac{{(Z_{\alpha/2})^2 \cdot (p \cdot (1 - p) + p \cdot (1 - p))}}{{E^2}} \]

Where:
- \( n \) = sample size in each group
- \( Z_{\alpha/2} \) = critical value for the desired level of confidence (for 99.9% confidence level, \( Z_{\alpha/2} = 3.291 \))
- \( p \) = estimated proportion (we usually take 0.5 for the maximum variability)
- \( E \) = margin of error

Given:
- \( E = 0.08 \)
- \( Z_{\alpha/2} = 3.291 \)

We need to estimate \( p \). Since we don't have any prior information, we'll use \( p = 0.5 \) for maximum variability, which leads to the largest possible sample size.

Substituting the values into the formula:

\[ n = \frac{{(3.291)^2 \cdot (0.5 \cdot 0.5 + 0.5 \cdot 0.5)}}{{0.08^2}} \]

\[ n = \frac{{(10.837281) \cdot (0.5)}}{{0.0064}} \]

\[ n \approx \frac{{5.418641}}{0.0064} \]

\[ n \approx 847.287 \]

Since you can't have a fraction of a person, you would need to round up to the nearest whole number. Therefore, you would need at least 848 people from each city to be sampled to estimate the proportion of people in favor of privatization with a 99.9% confidence level and a margin of error of 0.08.

To find the probability of earning a return greater than 15% using a normal distribution with a mean of 4% and a standard deviation of 6%, we can use the Z-score formula and then consult a standard normal distribution table (also known as a Z-table) or use statistical software.

The Z-score formula is:

\[ Z = \frac{{X - \mu}}{\sigma} \]

Where:
- \( X \) is the value we're interested in (15% return),
- \( \mu \) is the mean (4% return),
- \( \sigma \) is the standard deviation (6%).

\[ Z = \frac{{15 - 4}}{6} = \frac{{11}}{6} \]

\[ Z \approx 1.833 \]

Now, we look up the probability corresponding to a Z-score of 1.833 in a standard normal distribution table or use statistical software.

Using a standard normal distribution table, we find that the probability of a Z-score being greater than 1.833 is approximately 0.0336 or 3.36%.

So, the investor has approximately a 3.36% chance of earning a return greater than 15% if the investment returns are normally distributed.

The standard deviation of the sample mean, also known as the standard error (SE), can be calculated using the formula:

\[ SE = \frac{\sigma}{\sqrt{n}} \]

where:
- \(\sigma\) is the population standard deviation, and
- \(n\) is the sample size.

Given:
- The population standard deviation (\(\sigma\)) is 2 inches,
- The sample size (\(n\)) is 16.

Substituting the given values into the formula, we get:

\[ SE = \frac{2}{\sqrt{16}} \]

Since \(\sqrt{16} = 4\), the calculation becomes:

\[ SE = \frac{2}{4} = 0.5 \]

The standard deviation of the sample mean, or the standard error, is 0.5 inches.

Given that the site averages 12 damaged pairs for every 45,000 it produces, the probability of receiving a damaged pair of gloves (\(P_{damaged}\)) can be calculated as follows:

\[ P_{damaged} = \frac{\text{Number of damaged pairs}}{\text{Total pairs produced}} \]

\[ P_{damaged} = \frac{12}{45000} \]

Let's calculate \(P_{damaged}\):

\[ P_{damaged} = 0.0002666666666666667 \]

This probability represents the chance of receiving a damaged pair of gloves. To find the probability that all 20 pairs in Sean's order are damaged, we raise \(P_{damaged}\) to the 20th power, since the events are independent (the outcome of one pair being damaged does not affect the outcome of another pair being damaged):

\[ P_{all\ damaged} = (P_{damaged})^{20} \]

Substituting \(P_{damaged}\) into the equation:

\[ P_{all\ damaged} = (0.0002666666666666667)^{20} \]

Calculating \(P_{all\ damaged}\) gives us:

\[ P_{all\ damaged} \approx 3.31 \times 10^{-72} \]

Thus, the probability that all 20 pairs of gloves in Sean's order are damaged is extremely small, close to 0. This calculation shows the extreme unlikelihood of receiving an order where every pair of gloves is damaged, given the manufacturer's low rate of defects.

To calculate \(P(A)\), where \(A = \{q, t\}\), we need to sum the probabilities of the outcomes in event \(A\). Given that \(P(q) = 0.3\) and \(P(t) = 0.2\), the probability of \(A\) can be calculated as follows:

\[ P(A) = P(q) + P(t) \]

Substituting the given values:

\[ P(A) = 0.3 + 0.2 \]

The probability of event \(A\), \(P(A)\), is 0.5.

To determine the average days per week that the team plays softball, we can calculate the expected value (average) of the number of days they play. The expected value \(E(X)\) is given by the formula:

\[ E(X) = \sum (x_i \cdot P(x_i)) \]

where \(x_i\) represents the number of days the team plays in a week, and \(P(x_i)\) is the probability of playing \(x_i\) days.

Given:
- The probability they play 2 days a week is 0.3,
- The probability they play 3 days a week is 0.2,
- The probability they play 1 day a week is 0.5.

Let's calculate the expected value:

\[ E(X) = (2 \times 0.3) + (3 \times 0.2) + (1 \times 0.5) \]
\[ E(X) = 0.6 + 0.6 + 0.5 \]
\[ E(X) = 1.7 \]

Thus, the average days per week that the team plays softball is calculated to be 1.7 days.

To find the probability that the average height of a sample of 16 boys is 71 inches or taller, given a population with a mean height of 68 inches and a standard deviation of 2 inches, we can use the Z-score formula for a sample mean. The Z-score formula is:

\[ Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}} \]

where:
- \(\bar{X}\) is the sample mean,
- \(\mu\) is the population mean,
- \(\sigma\) is the population standard deviation, and
- \(n\) is the sample size.

Given:
- The population mean (\(\mu\)) is 68 inches,
- The population standard deviation (\(\sigma\)) is 2 inches,
- The sample size (\(n\)) is 16,
- The sample mean (\(\bar{X}\)) we're interested in is 71 inches.

We want to find the probability that the sample mean is 71 inches or more, so we calculate the Z-score using the formula:

\[ Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}} \]

Substituting the given values:

\[ Z = \frac{71 - 68}{\frac{2}{\sqrt{16}}} \]

\[ Z = \frac{3}{\frac{2}{4}} \]

\[ Z = \frac{3}{0.5} = 6 \]

The Z-score is 6, which means the sample mean is 6 standard deviations above the population mean. To find the probability associated with a Z-score of 6, we look at the right tail of the standard normal distribution, since we're interested in the probability of the sample mean being greater than or equal to 71 inches.

The probability associated with a Z-score of 6 is extremely small, as indicated by our calculation, which resulted in approximately \(9.87 \times 10^{-10}\). This probability value represents the area under the standard normal curve to the right of Z = 6, indicating that the likelihood of randomly selecting 16 boys with an average height of 71 inches or more is virtually zero.

To calculate the median of the sample heights, we first need to sort the heights in ascending order and then find the middle value (or the average of the two middle values if the sample size is even). Here's how it's done step-by-step with the given heights:

Given heights (in inches): {62.3, 68, 64, 66.1, 65.4, 61.8, 65.2, 68.6}

1. **Sort the heights in ascending order**:
- Sorted heights: {61.8, 62.3, 64, 65.2, 65.4, 66.1, 68, 68.6}

2. **Determine the sample size (n)**:
- The sample size (n) = 8 (even number)

3. **Find the middle position(s) since n is even**:
- Since n = 8, the middle positions are the 4th and 5th numbers in the sorted list.

4. **Identify the middle values**:
- The 4th number is 65.2, and the 5th number is 65.4.

5. **Calculate the median**:
- The median is the average of the 4th and 5th numbers: (65.2 + 65.4) / 2 = 65.3 inches.

Therefore, the median height of the sampled students is 65.3 inches. This value represents the midpoint of the height distribution in the sample, with half of the students having heights below this value and half having heights above it.

In statistics, the term **"population"** refers to the complete set of items or individuals that share one or more characteristics and are of interest for some question or experiment. A population can include people, objects, events, observations, or measurements. The defining feature of a population is that it encompasses all possible subjects that fit the criteria of the study or the question being asked, not just a subset.

For example:
- If you are studying the average height of adult men in a particular country, the population would include all adult men in that country.
- In a study examining the lifespan of a particular model of car, the population would be every car of that model.

Researchers often collect data from a **sample**, which is a subset of the population, to make inferences or predictions about the population because it's usually impractical or impossible to collect data from every member of the population. The goal is for the sample to be representative of the population, allowing conclusions drawn from the sample to be generalized back to the entire population.

To find the \(z\) value that satisfies \(P(Z \leq z) = 0.5636\) for a standard normal variable \(Z\), we look up the given probability in the standard normal distribution table or use a statistical software/tool that can invert the cumulative distribution function (CDF) of the standard normal distribution. This process is commonly referred to as finding the \(z\)-score or the percentile point.

The \(z\) value that satisfies the expression \(P(Z \leq z) = 0.5636\) for a standard normal variable \(Z\) is approximately \(0.16\).

To find \(P(o)\), we use the fact that the sum of all probabilities in a sample space must equal 1. We are given:

- \(P(l) = 0.3\)
- \(P(m) = 0.3\)
- \(P(n) = 0.2\)

To find \(P(o)\), we subtract the sum of the probabilities of \(l\), \(m\), and \(n\) from 1:

\[P(o) = 1 - (P(l) + P(m) + P(n))\]

Substituting the given values:

\[P(o) = 1 - (0.3 + 0.3 + 0.2)\]
\[P(o) = 1 - 0.8\]
\[P(o) = 0.2\]

Thus, \(P(o) = 0.2\), meaning the probability of outcome \(o\) is 0.2.

In the context of probability and statistics, when we discuss the **probability distribution** of a continuous random variable \(X\), we're focusing on how probabilities are allocated to various outcomes \(X\) can achieve within a continuous range. Unlike discrete random variables, which possess specific, countable outcomes, continuous random variables can assume any value within an interval, meaning their potential outcomes are uncountably infinite.

The **density function**, denoted as \(f(x)\), is crucial for defining the probability distribution of a continuous random variable. This function is characterized by several key features:

1. **Non-negativity**: \(f(x) \geq 0\) for all \(x\) within the domain of \(X\).
2. **Total area equals 1**: The integral of the density function \(f(x)\) over all possible values of \(X\) equals 1. This condition mirrors the concept that the probability of \(X\) assuming any value within its range is 100%.

To determine the probability that \(X\) lies within a certain interval \([a, b]\), we integrate the density function \(f(x)\) over this interval:

\[P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx\]

This calculation aggregates the "density" of \(X\) across the interval from \(a\) to \(b\), indicating the probability that \(X\) falls within that range. Given \(X\)'s continuity, the probability of \(X\) equalling any singular, precise value is \(0\). Probabilities are thus always computed over ranges of values.

In essence, the **probability distribution** of a continuous random variable, as outlined by its density function, offers a detailed map of how probabilities are distributed across different intervals of values that the variable can take on.

The normal distribution is described by the bell curve due to its characteristic shape, which resembles the outline of a bell. This shape arises from the mathematical properties of the normal distribution formula. Here's why the bell curve is a fitting description:

1. **Symmetry:** The normal distribution is perfectly symmetrical around its mean. This means that the left half of the distribution is a mirror image of the right half.

2. **Peak at the Mean:** The highest point of the curve occurs at the mean (\( \mu \)), which is also the median and mode in a normal distribution. This central peak tapers off symmetrically on both sides, decreasing as one moves away from the mean.

3. **Tails:** The curve approaches, but never touches, the horizontal axis. It extends infinitely in both directions, indicating that all values are possible for a normally distributed variable, but values far from the mean are less likely. These "tails" of the distribution decrease asymptotically towards the axis.

4. **Mathematical Equation:** The shape comes from the equation of the normal distribution. The equation involves \( e \) (Euler's number, approximately 2.71828) raised to the power of a negative quadratic function of the variable. This mathematical form ensures the symmetrical, bell-shaped curve.

The bell curve is a fundamental concept in statistics because it describes the expected distribution of many natural phenomena, from heights and test scores to errors in measurements, making it a powerful tool for statistical analysis and probability theory.

To find the sample mean, which is the average height in this case, you add up all the heights of the students in the sample and then divide that total by the number of students. Here's the simple formula:

\[ \text{Sample Mean} = \frac{\text{Sum of all heights}}{\text{Number of students}} \]

For our example, we added together the heights of the 5 students and divided by 5. This gave us the average height, or sample mean, of 65.3 inches.

The indicated value \(z_{0.02}\) refers to the z-score in a standard normal distribution where there is 0.02 (or 2%) of the area under the curve to the right of this z-score. This value is typically found using a standard normal (Z) table or statistical software, as it corresponds to the 98th percentile.

The indicated value of the standard normal variable \(Z\), \(z_{0.02}\), is approximately 2.05. This means that 2% of the area under the standard normal distribution curve lies to the right of \(z = 2.05\).

The goodness-of-fit chi-square (χ²) test is used to determine how well observed data fit an expected distribution. Larger values of χ² indicate a greater deviation from the expected distribution, suggesting that the observed data do not fit well. Therefore, the test is right-tailed because we are typically interested in whether the observed deviation is significantly greater than what would be expected by chance, indicating a poor fit.

A parameter is a numerical value that describes a specific characteristic of a population. In statistics, a population refers to the complete set of items that interest us, which could be people, objects, events, or measurements. Parameters are used to summarize this large set of data into a single value or a set of values that provide insights about the entire population.

Some common examples of parameters include:

- **Mean (Average):** The sum of all values divided by the number of values in the population.
- **Median:** The middle value when the data is arranged in order.
- **Mode:** The value that appears most frequently in the data set.
- **Standard Deviation:** A measure of the amount of variation or dispersion in the set of data values.

Unlike statistics, which are similar measures but derived from a sample of the population, parameters are considered fixed values since they describe the entire population. However, in practice, because it's often not feasible to study every member of a population, parameters are usually estimated through the use of samples and statistical methods.

The description provided, involving the summary of sample size, sample sizes in select subgroups, demographics like age, and the proportion of subjects of each breed, refers to activities in the realm of **Descriptive Statistics**. Descriptive statistics is the branch of statistics that deals with the organization, summarization, and visualization of data to describe the basic features of the data in a study. It provides simple summaries about the sample and the measures. Through descriptive statistics, researchers can present quantitative descriptions in a manageable form, involving measures of central tendency (like mean, median, mode), measures of variability (like range, variance, standard deviation), and frequency distributions, including proportions and percentages. This branch is crucial for the initial analysis of data, setting the stage for further statistical analysis or inferential statistics, where conclusions about the population are drawn from the sample data.

Given that \(A\) and \(B\) are independent events, the probability of both \(A\) and \(B\) occurring, denoted as \(P(A \cap B)\), can be calculated using the formula for the probability of the intersection of two independent events:

\[P(A \cap B) = P(A) \times P(B)\]

However, we are initially provided with \(P(A^c)\), the probability of the complement of \(A\), and \(P(B)\). Since \(P(A) + P(A^c) = 1\), we can find \(P(A)\) as:

\[P(A) = 1 - P(A^c)\]

Substituting \(P(A^c) = \frac{1}{5}\) into the equation gives us \(P(A)\):

\[P(A) = 1 - \frac{1}{5} = \frac{4}{5}\]

Now, using \(P(A) = \frac{4}{5}\) and the given \(P(B) = \frac{1}{8}\), we calculate \(P(A \cap B)\) as:

\[P(A \cap B) = \frac{4}{5} \times \frac{1}{8} = \frac{4}{40} = \frac{1}{10}\]

Therefore, the probability of both \(A\) and \(B\) occurring, \(P(A \cap B)\), is \(\frac{1}{10}\).

To construct a 99% confidence interval for the proportion of all students in the university who like to play football, given a random sample of 300 students where 75 declared their interest in playing football, we use the formula:

\[CI = \hat{p} \pm Z_{\frac{\alpha}{2}} \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\]

where:
- \(\hat{p}\) is the sample proportion, calculated as \(\hat{p} = \frac{75}{300}\),
- \(Z_{\frac{\alpha}{2}}\) is the Z-score for the 99% confidence level, which is approximately \(2.576\),
- \(n\) is the sample size, which is \(300\).

Substituting the given values into the formula, we get:

\[\hat{p} = \frac{75}{300} = 0.25\]

The confidence interval is calculated as:

\[CI = 0.25 \pm 2.576 \cdot \sqrt{\frac{0.25 \times (1 - 0.25)}{300}}\]

After the calculations, the confidence interval is found to be approximately:

\[CI \approx (0.186, 0.314)\]

This means we are 99% confident that the true proportion of all students in the university who like to play football is between 18.6% and 31.4%.

To determine the \(F_{0.025}\) test statistic value for degrees of freedom \(df_1 = 5\) and \(df_2 = 7\), we are referring to the F-distribution's critical value at the 0.025 significance level, with \(df_1\) and \(df_2\) representing the degrees of freedom for the numerator and denominator, respectively. This critical value is used in the context of hypothesis testing to decide whether to reject the null hypothesis, specifically in ANOVA or certain regression analyses where the F-test is applicable.

The \(F_{0.025}\) value is found in F-distribution tables or can be computed using statistical software. This value serves as a threshold: if the calculated F-statistic from your data exceeds this value, it indicates that there is a statistically significant difference between the groups being compared at the 0.025 significance level.

The \(F_{0.025}\) test statistic value for degrees of freedom \(df_1 = 5\) and \(df_2 = 7\) is approximately \(5.285\). This means that, at the 0.025 significance level, if your calculated F-statistic exceeds \(5.285\), there is a statistically significant difference between the groups you are comparing, leading to the rejection of the null hypothesis in the context of an ANOVA test or a similar statistical analysis.

To compute the sample median for the given data set \(\{-5, 4, 8, 5, 2, -1, 3, -6\}\), we first need to sort the data set in ascending order, which results in \(\{-6, -5, -1, 2, 3, 4, 5, 8\}\). The sorted data set has an even number of observations (\(N = 8\)).

For data sets with an even number of observations, the median is calculated as the average of the two middle numbers. In this case, the middle numbers are the fourth and fifth observations in the sorted list, which are \(2\) and \(3\), respectively.

Thus, the median is calculated as follows:

\[
\text{Median} = \frac{2 + 3}{2} = 2.5
\]

Therefore, the sample median of the given data set is \(2.5\).

To calculate the percentage of students who will be promoted to the next grade, we utilize the properties of the normal distribution. The normal distribution is defined by two parameters: the mean \(\mu\) and the standard deviation \(\sigma\). For this scenario, we have \(\mu = 60\) and \(\sigma = 5.5\), and the promotion cutoff score is \(55\).

The percentage of students scoring above the cutoff can be found by calculating the area under the normal distribution curve to the right of the cutoff score. This area represents the probability that a randomly selected student's score exceeds the cutoff. The cumulative distribution function (CDF) gives us the probability that a variable takes a value less than or equal to a certain value in a distribution. Therefore, to find the probability of scoring above the cutoff, we subtract the CDF at the cutoff score from 1:

\[P(X > 55) = 1 - P(X \leq 55) = 1 - F(55)\]

where \(F(55)\) is the CDF for a normal distribution with mean \(\mu = 60\) and standard deviation \(\sigma = 5.5\) evaluated at \(55\).

In mathematical terms, this calculation is:

\[P(X > 55) = 1 - \Phi\left(\frac{55 - \mu}{\sigma}\right)\]

where \(\Phi\) denotes the standard normal CDF. After computing this expression, we found that approximately \(81.83\%\) of the students will be promoted to the next grade, as this percentage of students scores above \(55\).

The term "population" refers to the complete set of all items that share at least one property of interest. In contrast, a "sample" is a subset of the population, usually selected for the purpose of observation and analysis. Based on these definitions:

**Heights of all students in a college** represents a population because it includes the entire group of interest (all students in the college) without exception.

To construct an 80% confidence interval for the mean time spent in the beauty salon by all the women in the city, we use the formula for a confidence interval for the mean, which is:

\[
\bar{x} \pm Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}
\]

where:
- \(\bar{x}\) is the sample mean,
- \(Z_{\alpha/2}\) is the Z-score corresponding to the desired confidence level (in this case, 80%),
- \(\sigma\) is the standard deviation of the sample, and
- \(n\) is the sample size.

Given:
- \(\bar{x} = 44\) minutes,
- \(\sigma = 16\) minutes, and
- \(n = 200\).

First, we need to find \(Z_{\alpha/2}\) for an 80% confidence level, which means \(\alpha = 1 - 0.80 = 0.20\), so \(\alpha/2 = 0.10\). We'll then calculate the confidence interval.

For an 80% confidence interval, the Z-score (\(Z_{\alpha/2}\)) is approximately \(1.282\). The margin of error is about \(1.45\) minutes.

Therefore, the 80% confidence interval for the mean time spent in the salon by all the women in the city is approximately \(42.55\) minutes to \(45.45\) minutes. This interval suggests that we can be 80% confident that the true mean time spent in the beauty salon by all women in the city lies within this range.

When constructing confidence intervals, especially for proportions, the length of the interval is influenced by the confidence level and the variability within the sample, but not directly by degrees of freedom, which are more relevant to t-distributions used for means with smaller sample sizes.
A higher confidence level means you want to be more certain about capturing the true parameter, which requires widening the interval. As confidence levels increase (e.g., from 90% to 95% to 99%), the corresponding Z-scores also increase, leading to a longer confidence interval.

A random experiment is defined as an experiment or a process for which the outcome cannot be predicted with certainty in advance. An experiment conducted to verify Archimedes' principle is not a random experiment. Archimedes' principle is a law of physics that describes the buoyant force on an object submerged in a fluid. Experiments to verify it are designed to observe this principle in action, rather than to obtain unpredictable outcomes. The results of such experiments, when conducted correctly, are expected and based on established physical laws, not on randomness.

A census is preferred over a sample when it's feasible to measure or observe every member of the population of interest, especially when the population is small or when accurate, comprehensive data is critical.

To find \(P(\hat{p} \leq 0.6)\), where \(\hat{p}\) is the sample proportion, we're interested in calculating the probability that the sample proportion is less than or equal to 0.6 given a population proportion \(p = 0.56\) and a sample size of \(n = 120\). To do this, we'll use the Z-score formula for a sample proportion, which is:

\[
Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1 - p)}{n}}}
\]

In this case, \(\hat{p} = 0.6\), \(p = 0.56\), and \(n = 120\). Let's calculate the Z-score and then use it to find the probability \(P(\hat{p} \leq 0.6)\).

The Z-score for a sample proportion of \(\hat{p} = 0.6\) is approximately \(0.883\), and the probability \(P(\hat{p} \leq 0.6)\) is approximately \(81.13\%\). This means there's an 81.13% chance that the sample proportion will be 0.6 or less, given a population proportion of 0.56 with a sample size of 120.

For a value \(x\) in a data set, if \(P\) percent of the values are less than or equal to \(x\), then \(x\) is called the **Pth percentile**.

Given:
- Number of damaged pairs, \(D = 12\)
- Total number of pairs produced, \(T = 45,000\)
- Number of pairs ordered, \(N = 20\)

First, find the probability of getting a damaged pair of gloves (\(P_{\text{Damaged}}\)):
\[ P_{\text{Damaged}} = \frac{D}{T} = \frac{12}{45,000} \]

Next, calculate the probability of getting a non-damaged (perfect) pair of gloves (\(P_{\text{Perfect}}\)):
\[ P_{\text{Perfect}} = 1 - P_{\text{Damaged}} \]

Finally, find the probability of all 20 pairs being perfect (\(P_{\text{All 20 Perfect}}\)):
\[ P_{\text{All 20 Perfect}} = (P_{\text{Perfect}})^N = (P_{\text{Perfect}})^{20} \]

Plugging in the values:
\[ P_{\text{Damaged}} = \frac{12}{45,000} \]

\[ P_{\text{Perfect}} = 1 - \frac{12}{45,000} \]

\[ P_{\text{All 20 Perfect}} = \left(1 - \frac{12}{45,000}\right)^{20} \]

Calculation result:
\[ P_{\text{All 20 Perfect}} \approx 0.9946801561844798 \]

Therefore, the probability that all 20 pairs are perfect is approximately 0.995.

The binomial probability formula, which calculates the probability of achieving exactly \(k\) successes (in this case, wins) in \(n\) independent Bernoulli trials, is expressed as:

\[ P(X = k) = {n \choose k} \cdot p^k \cdot (1-p)^{n-k} \]

Here:
- \(n\) is the number of trials (games played, which is 10),
- \(k\) is the number of successes (wins),
- \(p\) is the probability of success on a single trial (winning a game, which is 0.10),
- \(1-p\) is the probability of failure (not winning a game).

To calculate the probability of winning "at most once" over 10 games, we sum the probabilities of winning exactly 0 times and exactly 1 time:

1. **Winning 0 times** (\(k = 0\)):

\[ P(X = 0) = {10 \choose 0} \cdot 0.10^0 \cdot 0.90^{10} \]

2. **Winning 1 time** (\(k = 1\)):

\[ P(X = 1) = {10 \choose 1} \cdot 0.10^1 \cdot 0.90^9 \]

Thus, the total probability of winning at most once is:

\[ P(X \leq 1) = P(X = 0) + P(X = 1) \]

\[ P(X = 0) = 1 \cdot (0.90)^{10} \]

\[ P(X = 1) = 10 \cdot 0.10 \cdot (0.90)^9 \]

Therefore, the probability of Jeff winning at most once when playing the game 10 times is the sum of \( P(X = 0) \) and \( P(X = 1) \).

1. The probability of winning 0 times (\(P(X = 0)\)) is approximately \(0.349\).
2. The probability of winning exactly 1 time (\(P(X = 1)\)) is approximately \(0.387\).

Thus, the total probability of Jeff winning at most once (\(P(X \leq 1)\)) when playing the game 10 times is approximately \(0.736\), or 73.6%.

The most frequently occurring value in a sample data set is called the "mode."

The standard deviation of the sample mean, also known as the standard error of the mean, is smaller than the standard deviation of the population. Specifically, when the sample size is 2, the standard deviation of the sample mean is the standard deviation of the population divided by the square root of the sample size (in this case, the square root of 2). This relationship is expressed by the formula:

\[ \text{Standard Error of the Mean} = \frac{\sigma}{\sqrt{n}} \]

where \( \sigma \) is the population standard deviation and \( n \) is the sample size. So, with a sample size of 2:

\[ \text{Standard Error of the Mean} = \frac{\sigma}{\sqrt{2}} \]

Therefore, the standard deviation of the sample mean is \( \frac{1}{\sqrt{2}} \) times the standard deviation of the population when the sample size is 2.

The range of a data set is a measure of the spread or the dispersion of the values within the set. It's calculated by finding the difference between the largest and the smallest values in the set.

1. **Identify the Largest Value**: Look through the data set to find the maximum or largest value. In this case, the largest value is 8.

2. **Identify the Smallest Value**: Similarly, identify the minimum or smallest value in the data set. Here, the smallest value is 3.

3. **Calculate the Difference**: Subtract the smallest value from the largest value. This difference is the range of the data set. So, you calculate \(8 - 3 = 5\).

Thus, the range of the data set is 5. This value represents the total spread of values in the set, giving you a simple measure of variability within the data. The larger the range, the greater the difference between the smallest and largest values in the data set, indicating a wider dispersion of data points.

A random variable is called **continuous** if its possible values contain a whole interval of numbers.

Given the dataset \(D = \{1.5, 2.3, 1.7, 3.6, 2.9, 4.2\}\), we first sort it to get \(D_{\text{sorted}} = \{1.5, 1.7, 2.3, 2.9, 3.6, 4.2\}\).

To find the second quartile (Q2), which is the median:

- Since the number of observations in \(D_{\text{sorted}}\) is even (\(n = 6\)), Q2 is the average of the middle two values. Thus,
\[ Q2 = \frac{D_{\text{sorted}}[3] + D_{\text{sorted}}[4]}{2} \]
\[ Q2 = \frac{2.3 + 2.9}{2} \]
\[ Q2 = \frac{5.2}{2} \]
\[ Q2 = 2.6 \]

Therefore, the second quartile (Q2) of the dataset is \(2.6\).

To calculate the sample mean (\(\bar{x}\)) of the cholesterol levels for the selected sample:

Given the cholesterol levels of the 7 patients: \(128\), \(127\), \(153\), \(144\), \(132\), \(120\), \(115\),

The formula for the sample mean is:

\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \]

Where:
- \(x_i\) represents each value in the dataset,
- \(n\) is the number of observations in the dataset.

Substituting the given values into the formula:

\[ \bar{x} = \frac{128 + 127 + 153 + 144 + 132 + 120 + 115}{7} \]

\[ \bar{x} = \frac{919}{7} \]

\[ \bar{x} \approx 131.29 \]

Therefore, the sample mean of the cholesterol levels for the selected sample of 7 patients is approximately \(131.29\) mg/dL.

\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]

Given that the population standard deviation (\( \sigma \)) is \(2\) inches and the sample size (\( n \)) is \(100\), the calculation is:

\[ \sigma_{\bar{x}} = \frac{2}{\sqrt{100}} \]

\[ \sigma_{\bar{x}} = \frac{2}{10} \]

\[ \sigma_{\bar{x}} = 0.2 \]

Therefore, the standard deviation of the sample mean for this random sample of 100 male students is \(0.2\) inches.

Discrete random variables arise from a counting process.

Given a continuous random variable \( X \) that is normally distributed with a mean (\( \mu \)) of 50, and the probability \( P(X > 55) = 0.25 \), we can use the symmetry of the normal distribution to answer the query about \( P(X < 45) \).

Since the normal distribution is symmetric around its mean, the distribution of values below the mean mirrors the distribution of values above the mean. Specifically, the value 55 is 5 units above the mean of 50, and thus 45 is 5 units below the mean of 50.

Given that \( P(X > 55) = 0.25 \), the equivalent area on the left side of the mean (i.e., below 45) should have the same probability due to the symmetry of the normal distribution. Therefore, the probability \( P(X < 45) \) is also \( 0.25 \).

Among the options provided, the one that is not a random variable is:

b. Speed of light in a vacuum

The speed of light in a vacuum is a constant physical quantity, not a variable that changes and has different values in different situations. It is universally recognized and fixed at approximately 299,792 kilometers per second. All other options describe quantities that can vary and therefore are considered random variables.

To estimate the minimum sample size required to achieve a desired confidence interval width for the mean time a child takes to arrange a jigsaw puzzle, we use the formula related to the sample size estimation for a mean:

\[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \]

Where:
- \( n \) is the sample size.
- \( Z \) is the z-value corresponding to the desired confidence level (here, 90%).
- \( \sigma \) is the population standard deviation (here, estimated as 4.2 minutes from another class).
- \( E \) is the margin of error (the desired precision, here ±3 minutes).

For a 90% confidence interval, the critical z-value (since the normal distribution table or calculator provides the critical values) corresponds approximately to 1.645 for a one-sided interval of 95% (as it splits the remaining 10% into 5% on each tail in a two-tailed test). We're looking at half the interval width as the margin of error \( E \), so we use ±3 minutes.

Calculate the necessary sample size using these inputs.

To estimate the mean time a child takes to arrange a jigsaw puzzle to within 3 minutes with 90% confidence, the kindergarten teacher needs a minimum sample size of 6 children.

To construct an 80% confidence interval for the population mean when the sample size is \( n = 25 \), the sample mean is \( \bar{x} = 12 \), and the sample standard deviation is \( s = 5 \), we can use the following steps:

1. **Determine the Confidence Level and Critical Value**: The confidence level is 80%. We need to find the critical value from the t-distribution because we are estimating the mean from a sample and the sample size is relatively small.

2. **Calculate the Margin of Error**: The margin of error is computed using the formula:
\[
\text{Margin of Error} = t^* \times \frac{s}{\sqrt{n}}
\]
where \( t^* \) is the critical value from the t-distribution for \( df = n - 1 = 24 \) degrees of freedom and a two-tailed test at the 80% confidence level.

3. **Construct the Confidence Interval**: The confidence interval for the population mean \( \mu \) is then given by:
\[
\text{CI} = \left( \bar{x} - \text{Margin of Error}, \bar{x} + \text{Margin of Error} \right)
\]

Calculate this, beginning with finding the appropriate t-value for an 80% confidence level.

The t-critical value for an 80% confidence interval with 24 degrees of freedom is approximately 1.318. The margin of error is calculated as approximately 1.318.

Thus, the 80% confidence interval for the population mean is approximately \( (10.68, 13.32) \). This interval suggests that we can be 80% confident that the population mean lies within this range.

The observed significance of a test of hypotheses is the area of the tail of the distribution cut off by the **test statistic**. This area represents the probability of observing a test statistic as extreme as, or more extreme than, the one calculated from the sample data, assuming that the null hypothesis is true. This probability is commonly referred to as the p-value.

A fair six-sided die has numbers 1 through 6. The numbers less than three are 1 and 2. To find the probability of rolling a number less than three, we need to find the proportion of outcomes that satisfy this condition.

There are two outcomes (rolling a 1 or rolling a 2) that satisfy the condition, and there are a total of six possible outcomes when rolling a fair die.

So, the probability of rolling a number less than three is:

\[ P(\text{number less than three}) = \frac{\text{Number of outcomes less than three}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3} \]

Thus, the probability of rolling a number less than three is \( \frac{1}{3} \) or approximately 0.3333.

In probability notation, \( P(G | H) \) represents the conditional probability of event G occurring given that event H has already occurred. Since events G and H are mutually exclusive, it means that they cannot occur at the same time. Therefore, if event H has occurred, event G cannot occur, and the probability of event G occurring given that event H has occurred is 0.

So, \( P(G | H) = 0 \).

The correct option is: "between 0 and 1."

In hypothesis testing concerning a single population proportion, the null hypothesis (\( H_0 \)) typically assumes that the population proportion (\( p \)) is equal to a specific value \( p_0 \), where \( p_0 \) is a number between 0 and 1.

To estimate the minimum sample size required to estimate the mean serving time with a 99% confidence level and a margin of error of 55 seconds, we can use the formula for sample size calculation:

\[ n = \left( \frac{{Z \cdot \sigma}}{{E}} \right)^2 \]

Where:
- \( n \) = sample size
- \( Z \) = Z-score corresponding to the desired confidence level (99% confidence level corresponds to a Z-score of approximately 2.576)
- \( \sigma \) = standard deviation of the population
- \( E \) = margin of error

Given:
- \( Z \) = 2.576 (for 99% confidence level)
- \( \sigma \) = 2.5 minutes = 150 seconds
- \( E \) = 55 seconds

Substitute the values into the formula:

\[ n = \left( \frac{{2.576 \cdot 150}}{{55}} \right)^2 \]

\[ n = \left( \frac{{386}}{{55}} \right)^2 \]

\[ n \approx \left( 7.018 \right)^2 \]

\[ n \approx 49.318 \]

Since the sample size must be a whole number, we round up to the nearest whole number:

\[ n \approx 50 \]

So, a minimum sample size of 50 is required to estimate the mean serving time with a 99% confidence level and a margin of error of 55 seconds.

Frequency is the number of times a value x appears in a data set.

A **sample** is any subset or subcollection of the population.

To find the probability that a random variable \( X \), uniformly distributed on the interval [0, 1], assumes a value greater than 0.90, we can use the properties of the uniform distribution.

The probability density function \( f(x) \) of a uniform distribution on the interval [a, b] (in this case, [0, 1]) is defined as:
\[
f(x) = \frac{1}{b - a}
\]
For the interval [0, 1], this simplifies to:
\[
f(x) = 1 \text{ for } x \in [0, 1]
\]

The probability that \( X \) is greater than a particular value \( c \) within [0, 1] is calculated as the length of the interval from \( c \) to 1. Specifically, for \( X > 0.90 \):
\[
P(X > 0.90) = 1 - 0.90 = 0.10
\]

Hence, the probability that \( X \) assumes a value greater than 0.90 is 0.10, or 10%.

Calculating \( \frac{4}{6} \times 100 \) yields approximately 66.67%. Thus, if we're directly using the position in the dataset where the total number of elements equals 6 and 2.9 is the fourth element, 2.9 is at the 66.67th percentile. This method is straightforward, counting the index directly against the total number of elements.

When two balanced dice are rolled, the total number of possible outcomes is \(6 \times 6 = 36\), since each die has 6 faces.

To find the probability of not rolling doubles, we first calculate the probability of rolling doubles. Rolling doubles happens when both dice show the same number, which can occur in the following combinations: \( (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \). There are 6 such outcomes.

Thus, the probability of rolling doubles is:
\[
\frac{6}{36} = \frac{1}{6}
\]

To find the probability of not rolling doubles, we subtract the probability of rolling doubles from 1:
\[
1 - \frac{1}{6} = \frac{5}{6}
\]

Hence, the probability of not rolling doubles is \(\frac{5}{6}\).

As the sample size of a random sample increases, the standard deviation of the sample mean **decreases**. This concept is reflected in the formula for the standard error of the mean, which is the standard deviation of the sample mean:

\[
\text{Standard Error} = \frac{\sigma}{\sqrt{n}}
\]

where \( \sigma \) is the population standard deviation and \( n \) is the sample size. As \( n \) increases, \( \sqrt{n} \) increases, making the denominator larger and thus reducing the standard error.

A **measurement** is a number or attribute computed for each member of a population or of a sample.

Let's calculate the probability using the z-score and the standard normal distribution table.

Given:
- Mean (\(\mu\)) = 15 hours
- Standard deviation (\(\sigma\)) = 3 hours

We want to find the probability that Noel plays video games between 15 and 18 hours a week.

First, calculate the z-score for 18 hours:

\[ z = \frac{18 - 15}{3} = \frac{3}{3} = 1 \]

Now we need to find the probability corresponding to a z-score of 1. Using the standard normal distribution table, the cumulative probability for a z-score of 1 is approximately 0.8413.

This value represents the probability that Noel plays video games up to 18 hours a week. Since the mean (15 hours) is the center of the distribution, the cumulative probability up to the mean (z = 0) is 0.5.

To find the probability that Noel plays video games between 15 and 18 hours, subtract the cumulative probability at the mean from the cumulative probability at the z-score of 1:

\[ P(15 < X < 18) = P(z < 1) - P(z < 0) \]
\[ P(15 < X < 18) = 0.8413 - 0.5 = 0.3413 \]

Thus, the probability that Noel plays video games between 15 and 18 hours a week is approximately 34.13%.

In this context, the sample consists of the 1,200 people who were randomly selected to ascertain their income. These individuals represent a subset of the total population of employed people in Omaha, Nebraska, and their incomes are used to estimate the average income of the entire population.

A simple linear regression model is a sum of two parts:

1. **Deterministic Part (Predicted Value)**: This is the linear component that represents the relationship between the independent variable \(x\) and the dependent variable \(y\). It is expressed as \( \hat{y} = \beta_0 + \beta_1 x \), where:
- \( \beta_0 \) is the y-intercept.
- \( \beta_1 \) is the slope of the regression line.
- \( x \) is the independent variable.
- \( \hat{y} \) is the predicted value of the dependent variable.

2. **Stochastic Part (Error Term or Residual)**: This is the random component that accounts for the variability in the dependent variable \(y\) that cannot be explained by the linear relationship with \(x\). It is represented as \( \epsilon \).

Combining these two parts, the simple linear regression model can be written as:

\[ y = \beta_0 + \beta_1 x + \epsilon \]

Here, \(y\) is the actual observed value of the dependent variable, and \(\epsilon\) is the error term that captures the deviation of the observed value from the predicted value.

As the sample size of a random sample increases, the standard deviation of the sample mean decreases.

This phenomenon occurs because the standard deviation of the sample mean, often referred to as the standard error of the mean, is calculated as the population standard deviation divided by the square root of the sample size (\(n\)). Mathematically, this relationship is expressed as:

\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} \]

where \(\sigma\) is the population standard deviation and \(n\) is the sample size. As \(n\) increases, the denominator increases, resulting in a smaller standard error. This reduction in the standard error signifies that the sample mean becomes a more precise estimate of the population mean as the sample size grows.

To construct a 99.9% confidence interval for the mean difference between the marks scored last week and those scored this week by all the students, we'll use the following formula for the confidence interval of the mean:

\[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{s}{\sqrt{n}} \right) \]

Where:
- \(\bar{x}\) is the sample mean
- \(z\) is the z-score corresponding to the desired confidence level
- \(s\) is the sample standard deviation
- \(n\) is the sample size

Given:
- Sample mean difference, \(\bar{x} = 12\)
- Sample standard deviation, \(s = 3\)
- Sample size, \(n = 35\)
- Confidence level, \(99.9\%\)

For a 99.9% confidence level, the z-score (from the standard normal distribution table) is approximately \(z = 3.291\).

Now, calculate the standard error:

\[ \text{Standard Error} = \frac{s}{\sqrt{n}} = \frac{3}{\sqrt{35}} \approx \frac{3}{5.92} \approx 0.507 \]

Next, calculate the margin of error:

\[ \text{Margin of Error} = z \times \text{Standard Error} = 3.291 \times 0.507 \approx 1.669 \]

Finally, construct the confidence interval:

\[ \text{Lower Limit} = \bar{x} - \text{Margin of Error} = 12 - 1.669 = 10.331 \]
\[ \text{Upper Limit} = \bar{x} + \text{Margin of Error} = 12 + 1.669 = 13.669 \]

Therefore, the 99.9% confidence interval for the mean difference between the marks scored last week and those scored this week by all the students is approximately \((10.331, 13.669)\).

The parameter \( \text{df}_2 \) in the F-distribution is often referred to as the **degrees of freedom for the denominator**.

In the context of an F-distribution, there are two sets of degrees of freedom:
- \( \text{df}_1 \): the degrees of freedom for the numerator (associated with the variance estimate in the numerator of the F-ratio).
- \( \text{df}_2 \): the degrees of freedom for the denominator (associated with the variance estimate in the denominator of the F-ratio).

The F-distribution is typically used in the context of ANOVA (Analysis of Variance) and regression analysis to compare variances and test hypotheses about the equality of multiple group means.

To find the probability that the battery life for a randomly selected cell phone is below 3 hours, given that the battery life is normally distributed with a mean (\(\mu\)) of 4 hours and a standard deviation (\(\sigma\)) of 0.8 hours, we need to calculate the corresponding z-score and then find the probability associated with that z-score from the standard normal distribution.

The z-score is calculated using the formula:
\[
z = \frac{X - \mu}{\sigma}
\]
where:
- \( X \) is the value of interest (3 hours),
- \( \mu \) is the mean (4 hours),
- \( \sigma \) is the standard deviation (0.8 hours).

Plugging in the values, we get:
\[
z = \frac{3 - 4}{0.8} = \frac{-1}{0.8} = -1.25
\]

Next, we look up the z-score of -1.25 in the standard normal distribution table or use a calculator to find the probability that a z-score is less than -1.25.

The cumulative probability for \( Z = -1.25 \) is approximately 0.1056.

Therefore, the probability that the battery life for a randomly selected cell phone is below 3 hours is approximately **0.1056** or **10.56%**.

For a sample proportion \( p \) to have an approximately normal distribution, one of the conditions is that the sample size \( n \) is large enough such that both \( np \) and \( n(1 - p) \) are greater than 5. This ensures that the distribution of the sample proportion can be approximated by a normal distribution.

The condition above involves the range of plausible values for the sample proportion \( p \). Specifically, we want to ensure that the interval \([p - 3\sigma_p, p + 3\sigma_p]\) lies wholly within the interval \([0, 1]\). Here, \( \sigma_p \) is the standard error of the sample proportion, given by:

\[
\sigma_p = \sqrt{\frac{p(1 - p)}{n}}
\]

This condition ensures that the sample proportion remains within the valid probability range of 0 to 1.

The interval \([p - 3\sigma_p, p + 3\sigma_p]\) should lie wholly within the interval **[0, 1]**.

In other words, for the sample proportion \( p \) to be approximately normal, the interval:

\[
[p - 3\sigma_p, p + 3\sigma_p]
\]

must be entirely contained within the interval:

\[
[0, 1]
\]

To determine the probability that a randomly selected student spends more than $300 for textbooks each semester, we need to use the properties of the normal distribution.

Given:
- Mean (\(\mu\)): $250
- Standard deviation (\(\sigma\)): $50
- Amount (\(X\)): $300

We first need to convert the given amount to a z-score using the formula:
\[
z = \frac{X - \mu}{\sigma}
\]

Plugging in the values:
\[
z = \frac{300 - 250}{50} = \frac{50}{50} = 1
\]

Next, we look up the z-score of 1 in the standard normal distribution table to find the probability that \(Z \leq 1\).

The cumulative probability for \(Z = 1\) is approximately 0.8413. This value represents the probability that a randomly selected student spends less than $300 for textbooks.

To find the probability that a student spends more than $300, we need to subtract this cumulative probability from 1:
\[
P(X > 300) = 1 - P(X \leq 300) = 1 - 0.8413 = 0.1587
\]

Therefore, the probability that a randomly selected student spends more than $300 for textbooks each semester is approximately **0.1587** or **15.87%**.

To estimate the mean time that employees take to complete an assigned project with 99.8% confidence to within 1 day, we need to determine the minimum sample size required.

Given:
- Confidence level: 99.8%
- Margin of error (\(E\)): 1 day
- Standard deviation (\(\sigma\)): 3 days

We use the formula for the sample size \(n\) for estimating a mean:
\[
n = \left( \frac{Z_{\alpha/2} \cdot \sigma}{E} \right)^2
\]

First, we need to determine the critical value \(Z_{\alpha/2}\) for a 99.8% confidence level. The confidence level of 99.8% means that \(\alpha = 1 - 0.998 = 0.002\). Thus, \(\alpha/2 = 0.001\).

We find \(Z_{\alpha/2}\) using the standard normal distribution table or a calculator:
\[
Z_{\alpha/2} \approx 3.090
\]

Next, we plug in the values into the formula:
\[
n = \left( \frac{3.090 \cdot 3}{1} \right)^2
\]

Simplify the expression:
\[
n = \left( \frac{9.27}{1} \right)^2 = 9.27^2 = 85.9329
\]

Since the sample size must be a whole number, we round up to the nearest whole number:
\[
n \approx 86
\]

Therefore, the minimum sample size required to estimate the mean time that employees take to complete an assigned project with 99.8% confidence to within 1 day is **86**.

To find \( P(-2 < Z < 2) \) for a standard normal variable \( Z \), we use the standard normal distribution table (z-table) or a standard normal distribution calculator.

The cumulative probability \( P(Z < 2) \) for a standard normal variable is approximately 0.9772.

The cumulative probability \( P(Z < -2) \) for a standard normal variable is approximately 0.0228.

Therefore, the probability that \( Z \) falls between -2 and 2 is:

\[
P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) = 0.9772 - 0.0228 = 0.9544
\]

So, \( P(-2 < Z < 2) \) is approximately 0.9544 or 95.44%.

Descriptive statistics is the branch of statistics that involves organizing, displaying, and describing data.

To calculate the standard deviation of the population given the sample mean, sample size, and the standard deviation of the sample mean, we can use the relationship between the sample standard deviation (also known as the standard error of the mean) and the population standard deviation.

The formula for the standard error of the mean (\(\sigma_{\bar{x}}\)) is:

\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]

where \(\sigma\) is the population standard deviation and \(n\) is the sample size.

We are given:
- \(\sigma_{\bar{x}} = 20\)
- \(n = 400\)

We need to find \(\sigma\). Rearranging the formula to solve for \(\sigma\), we get:

\[
\sigma = \sigma_{\bar{x}} \times \sqrt{n}
\]

Substituting the given values:

\[
\sigma = 20 \times \sqrt{400}
\]

Since \(\sqrt{400} = 20\), we have:

\[
\sigma = 20 \times 20 = 400
\]

Therefore, the standard deviation of the population is 400.

In this context, the statistic is the proportion of people in the sample who are above the age of 80. The researcher calculated this proportion as follows:

\[
\hat{p} = \frac{120}{2400} = 0.05
\]

So, the value of the statistic is 0.05, or 5%. This represents the sample proportion of people above the age of 80 in the researcher's sample.

Qualitative data are measurements for which there is no natural numerical scale, but which consist of attributes, labels, or other nonnumerical characteristics.