Statistics Quiz with Solutions (5 Questions)

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.