Statistics Quiz with Solutions (5 Questions)

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.

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 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.

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 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.