Question

The length of time taken on the SAT for a group of students is normally distributed with 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. Find the probability that the sample mean is between two hours and three hours.

To find percentiles for means on the calculator, follow these steps.

2nd DIStR
3:invNorm

k = invNorm(area to the left of k,mean, standarddeviation/
`√(samplesize)`

where:

k = the kth percentile
mean is the mean of the original distribution
standard deviation is the standard deviation of the original distribution
sample size = n

Answer 1

Using the z-scores formula and the invNorm function, calculate z-scores for the lower and upper limits, find corresponding percentiles, and subtract to determine the probability that the sample mean is between 2 and 3 hours.

To find the probability that the sample mean is between two hours and three hours, we need to find the z-scores corresponding to these values and then use the normal distribution table.

Given:

- Mean
`(\(\mu\))` : 2.5 hours

- Standard Deviation
`(\(\sigma\))` : 0.25 hours

- Sample Size (\(n\)): 60

- Lower Limit: 2 hours

- Upper Limit: 3 hours

First, find the z-scores for both limits using the formula:

`\[ z = \frac{{X - \mu}}{{(\sigma)/(√(n))}} \]`

For the lower limit (2 hours):

`\[ z_{\text{lower}} = \frac{{2 - 2.5}}{{(0.25)/(√(60))}} \]`

For the upper limit (3 hours):

`\[ z_{\text{upper}} = \frac{{3 - 2.5}}{{(0.25)/(√(60))}} \]`

Now, use these z-scores to find the corresponding percentiles using the invNorm function on the calculator:

`\[ P(\text{Lower Limit}) = \text{invNorm}(z_{\text{lower}}, 0, 1) \]`

`\[ P(\text{Upper Limit}) = \text{invNorm}(z_{\text{upper}}, 0, 1) \]`

Finally, the probability that the sample mean is between two hours and three hours is given by the difference of these percentiles:

`\[ P(2 < \text{Sample Mean} < 3) = P(\text{Upper Limit}) - P(\text{Lower Limit}) \]`