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 41.0 minutes and a variance of 3.4 minutes. What is the probability, rounded to four decimal places, that a given student will complete the test in more than 35 minutes but less than 43 minutes?

Answer 1

Answer:

0.6832 = 68.32% probability that a given student will complete the test in more than 35 minutes but less than 43 minutes

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 41.0 minutes and a variance of 3.4 minutes.

This means that $\mu = 41, \sigma = 3.4$

What is the probability, rounded to four decimal places, that a given student will complete the test in more than 35 minutes but less than 43 minutes?

This is the p-value of Z when X = 43 subtracted by the p-value of Z when X = 35.

X = 43

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

$Z = \frac{43 - 41}{3.4}$

$Z = 0.59$

$Z = 0.59$ has a p-value of 0.7224

X = 35

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

$Z = \frac{35 - 41}{3.4}$

$Z = -1.76$

$Z = -1.76$ has a p-value of 0.0392

0.7224 - 0.0392 = 0.6832

0.6832 = 68.32% probability that a given student will complete the test in more than 35 minutes but less than 43 minutes