Statistics Quiz with Solutions (5 Questions)

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 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%.

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

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.