Question

the length of time taken on the sat for a group of students has a mean of 2.5 hours and a standard deviation of 0.25 hours. a sample size of n = 60 is drawn randomly from the population. a. what are the mean and standard deviation for the sample mean times to take sat tests? b. what does the distribution shape look like? why? c. find the probability that the sample mean is less than three hours. d. find the 95th percentile for the sample mean age (to one decimal place). interpret it. e. for part c) and d), is the assumption of normal necessary? explain.

Explanation:

Step1: Mean of sample mean

The mean of the sample - mean $\mu_{\bar{x}}$ is equal to the population mean $\mu$. Given $\mu = 2.5$ hours, so $\mu_{\bar{x}}=2.5$ hours.

Step2: Standard deviation of sample mean

The standard deviation of the sample - mean (also known as the standard error) is given by the formula $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. Given $\sigma = 0.25$ hours and $n = 60$, then $\sigma_{\bar{x}}=\frac{0.25}{\sqrt{60}}\approx0.032$ hours.

Step3: Distribution shape

The distribution of the sample mean is approximately normal. This is due to the Central Limit Theorem, which states that for a sample of size $n$ (where $n\geq30$), the sampling distribution of the sample mean $\bar{X}$ is approximately normal, regardless of the shape of the population distribution. Here, $n = 60\geq30$.

Step4: Probability that sample mean is less than 3 hours

First, calculate the z - score using the formula $z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}$. Substitute $\bar{x} = 3$, $\mu_{\bar{x}}=2.5$, and $\sigma_{\bar{x}}\approx0.032$ into the formula: $z=\frac{3 - 2.5}{0.032}=\frac{0.5}{0.032}\approx15.625$. Then, $P(\bar{X}<3)=P(Z < 15.625)\approx1$.

Step5: 95th percentile of sample mean

We know that if $Z$ is a standard normal random variable, the z - score corresponding to the 95th percentile, $z_{0.95}\approx1.645$. Using the formula $\bar{x}=\mu_{\bar{x}}+z\sigma_{\bar{x}}$, substitute $\mu_{\bar{x}} = 2.5$, $z = 1.645$, and $\sigma_{\bar{x}}\approx0.032$: $\bar{x}=2.5+1.645\times0.032=2.5 + 0.05264\approx2.6$ hours. Interpretation: 95% of the sample means of the time taken on the SAT are less than 2.6 hours.

Step6: Assumption of normality

For part c), since the sample size $n = 60\geq30$, by the Central Limit Theorem, the sampling distribution of the sample mean is approximately normal, so the assumption of normality is reasonable. For part d), to find the percentile, we use the properties of the normal distribution. Since the sample size is large ($n = 60$), the Central Limit Theorem allows us to assume that the sampling distribution of the sample mean is approximately normal, so the assumption of normality is necessary and reasonable.

Answer:

a. Mean: 2.5 hours, Standard deviation: 0.032 hours
b. Approximately normal. Because of the Central Limit Theorem ($n = 60\geq30$).
c. Approximately 1
d. 2.6 hours. 95% of the sample means of the time taken on the SAT are less than 2.6 hours.
e. Yes. For both c) and d), since $n = 60\geq30$, the Central Limit Theorem allows us to assume the sampling distribution of the sample mean is approximately normal.