Question

Suppose the length of telephone calls form a normal distribution with a mean length of 8.0 min and a standard deviation of 2.5 min. The probability that telephone call selected a random will last more than 15.5 min is most nearly
(A) 0.0013
(B) 0.0026
(C) 0.2600
(D) 0.9987

Answer 1

Answer:

(A) 0.0013

Step-by-step explanation:

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

In a set with mean $\mu$ and standard deviation $\sigma$, 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:

$\mu = 8, \sigma = 2.5$

The probability that telephone call selected a random will last more than 15.5 min is most nearly

This is 1 subtracted by the pvalue of Z when X = 15.5.

So

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

$Z = \frac{15.5 - 8}{2.5}$

$Z = 3$

$Z = 3$ has a pvalue of 0.9987

1 - 0.9987 = 0.0013

So the correct answer is:

(A) 0.0013