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2 answers:
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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:
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:
(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
has a pvalue of 0.9332
X = 3.6
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
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
has a pvalue of 0.9998
X = 4.2
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
has a pvalue of 0.9998
X = 3
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.
They last at least 4.3 minutes