Answer:
a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes
b) 0.0668 = 6.68% of the calls last more than 4.2 minutes
c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes
d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes
e) They last at least 4.3 minutes
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 = 3.6, \sigma = 0.4](https://tex.z-dn.net/?f=%5Cmu%20%3D%203.6%2C%20%5Csigma%20%3D%200.4)
(a) What fraction of the calls last between 3.6 and 4.2 minutes?
This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.
X = 4.2
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{4.2 - 3.6}{0.4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B4.2%20-%203.6%7D%7B0.4%7D)
![Z = 1.5](https://tex.z-dn.net/?f=Z%20%3D%201.5)
has a pvalue of 0.9332
X = 3.6
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{3.6 - 3.6}{0.4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B3.6%20-%203.6%7D%7B0.4%7D)
![Z = 0](https://tex.z-dn.net/?f=Z%20%3D%200)
has a pvalue of 0.5
0.9332 - 0.5 = 0.4332
0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes
(b) What fraction of the calls last more than 4.2 minutes?
This is 1 subtracted by the pvalue of Z when X = 4.2. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{4.2 - 3.6}{0.4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B4.2%20-%203.6%7D%7B0.4%7D)
![Z = 1.5](https://tex.z-dn.net/?f=Z%20%3D%201.5)
has a pvalue of 0.9332
1 - 0.9332 = 0.0668
0.0668 = 6.68% of the calls last more than 4.2 minutes
(c) What fraction of the calls last between 4.2 and 5 minutes?
This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So
X = 5
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{5 - 3.6}{0.4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B5%20-%203.6%7D%7B0.4%7D)
![Z = 3.5](https://tex.z-dn.net/?f=Z%20%3D%203.5)
has a pvalue of 0.9998
X = 4.2
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{4.2 - 3.6}{0.4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B4.2%20-%203.6%7D%7B0.4%7D)
![Z = 1.5](https://tex.z-dn.net/?f=Z%20%3D%201.5)
has a pvalue of 0.9332
0.9998 - 0.9332 = 0.0666
0.0666 = 6.66% of the calls last between 4.2 and 5 minutes
(d) What fraction of the calls last between 3 and 5 minutes?
This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.
X = 5
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{5 - 3.6}{0.4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B5%20-%203.6%7D%7B0.4%7D)
![Z = 3.5](https://tex.z-dn.net/?f=Z%20%3D%203.5)
has a pvalue of 0.9998
X = 3
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{3 - 3.6}{0.4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B3%20-%203.6%7D%7B0.4%7D)
![Z = -1.5](https://tex.z-dn.net/?f=Z%20%3D%20-1.5)
has a pvalue of 0.0668
0.9998 - 0.0668 = 0.9330
0.9330 = 93.30% of the calls last between 3 and 5 minutes
(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?
At least X minutes
X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![1.75 = \frac{X - 3.6}{0.4}](https://tex.z-dn.net/?f=1.75%20%3D%20%5Cfrac%7BX%20-%203.6%7D%7B0.4%7D)
![X - 3.6 = 0.4*1.75](https://tex.z-dn.net/?f=X%20-%203.6%20%3D%200.4%2A1.75)
![X = 4.3](https://tex.z-dn.net/?f=X%20%3D%204.3)
They last at least 4.3 minutes