Question

We know that the length of time required for a student to complete a particular aptitude test has a normal distribution with a mean of 40 minutes and a variance of 4 minutes. What is the probability that a given student will complete the test in more than 36 minutes but less than 44 minutes?

Answer 1

Answer:

95.44% probability that a given student will complete the test in more than 36 minutes but less than 44 minutes

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation(which is the square root of the variance) $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 40, \sigma = \sqrt{4} = 2$

What is the probability that a given student will complete the test in more than 36 minutes but less than 44 minutes?

This is the pvalue of Z when X = 44 subtracted by the pvalue of Z when X = 36. So

X = 44

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{44 - 40}{2}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

X = 36

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{36 - 40}{2}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

0.9772 - 0.0228

0.9544

95.44% probability that a given student will complete the test in more than 36 minutes but less than 44 minutes