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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?

Respuesta :

Notice that:

[tex]P(x<10)=P(x>14)_{}=1-P(x\leq14)[/tex]

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

[tex]z=\frac{(14-12)}{2}=1[/tex]

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

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