Statistics Quiz with Solutions (10 Questions)

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 \( 0.01\% \).

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.

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 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.

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 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%.

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%.

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.

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}.

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.