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Sunny_sXe [5.5K]

1 year ago

11

Assume a normal distribution and that the average phone call in a certain town lasted 12 min with a standard deviation of 2 min

What percentage of the calls lasted lessthan 10 min?

1 answer:

fredd [130]1 year ago

6 0

Notice that:

$P(x14)_{}=1-P(x\leq14)$

Now, we use the formula for the z-score:

$z=\frac{(14-12)}{2}=1$

Then the percentage of the calls that lasted less than 10 min is 15.87%

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