Questions: One college class had a total of 70 students. The average score for the class on the last exam was 83.4 with a standard deviation of 5.2. A random sample of 39 students was selected. a. Calculate the standard error of the mean b. What is the probability that the sample mean will be less than 85 ? c. What is the probability that the sample mean will be more than 84 ? d. What is the probability that the sample mean will be between 82.5 and 84.5 ? a. The standard error of the mean is (Round to two decimal places as needed.)

One college class had a total of 70 students. The average score for the class on the last exam was 83.4 with a standard deviation of 5.2. A random sample of 39 students was selected.

a. Calculate the standard error of the mean

b. What is the probability that the sample mean will be less than 85 ?

c. What is the probability that the sample mean will be more than 84 ?

d. What is the probability that the sample mean will be between 82.5 and 84.5 ?

a. The standard error of the mean is (Round to two decimal places as needed.)

Transcript text: One college class had a total of 70 students. The average score for the class on the last exam was 83.4 with a standard deviation of 5.2. A random sample of 39 students was selected. a. Calculate the standard error of the mean b. What is the probability that the sample mean will be less than 85 ? c. What is the probability that the sample mean will be more than 84 ? d. What is the probability that the sample mean will be between 82.5 and 84.5 ? a. The standard error of the mean is $\square$ (Round to two decimal places as needed.)

Solution

Solution Steps

To solve the given problems, we need to use concepts from statistics, specifically the standard error of the mean and properties of the normal distribution.

a. The standard error of the mean (SEM) is calculated using the formula: \[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \] where $\sigma$ is the standard deviation and $n$ is the sample size.

b. To find the probability that the sample mean will be less than 85, we need to calculate the z-score and then use the cumulative distribution function (CDF) of the standard normal distribution.

c. Similarly, to find the probability that the sample mean will be more than 84, we calculate the z-score and use the CDF.

Solution Approach

Step 1: Calculate the Standard Error of the Mean

The standard error of the mean (SEM) is calculated using the formula: \[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \] where $\sigma = 5.2$ and $n = 39$.

\[ \text{SEM} = \frac{5.2}{\sqrt{39}} \approx 0.8327 \]

Step 2: Calculate the Probability that the Sample Mean will be Less than 85

To find the probability that the sample mean will be less than 85, we first calculate the z-score: \[ z = \frac{\bar{x} - \mu}{\text{SEM}} \] where $\bar{x} = 85$, $\mu = 83.4$, and $\text{SEM} = 0.8327$.

\[ z = \frac{85 - 83.4}{0.8327} \approx 1.9215 \]

Using the cumulative distribution function (CDF) of the standard normal distribution: \[ P(\bar{x} < 85) = \Phi(1.9215) \approx 0.9727 \]

Step 3: Calculate the Probability that the Sample Mean will be More than 84

To find the probability that the sample mean will be more than 84, we calculate the z-score: \[ z = \frac{\bar{x} - \mu}{\text{SEM}} \] where $\bar{x} = 84$, $\mu = 83.4$, and $\text{SEM} = 0.8327$.

\[ z = \frac{84 - 83.4}{0.8327} \approx 0.7206 \]

Using the cumulative distribution function (CDF) of the standard normal distribution: \[ P(\bar{x} > 84) = 1 - \Phi(0.7206) \approx 0.2356 \]