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Finger [1]
1 year ago
8

Assume a normal distribution and that the average phone call in a certain town lasted 4 min, with a standard deviation of 1 min.

What percentage of the calls lasted lessthan 3 min?
Mathematics
1 answer:
olya-2409 [2.1K]1 year ago
4 0

SOLUTION

The mean is 4min

standard deviation is 1min

the z score is

z=\frac{x-\bar{x}}{\sigma}

where

\begin{gathered} x=3 \\ \bar{x}=4 \\ \sigma=1 \end{gathered}

then we have

\begin{gathered} z=\frac{3-4}{1}=\frac{-1}{1}=-1 \\ z=-1 \end{gathered}

The probability the call lasted less than 3 min will be

Therefore, the probability that (z < -1 ) is

[tex]\begin{gathered} Pr(z<-1)=Pr(0Hence, the percentage of the calls that lasted less than 3 min is 16%

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