Question

The lengths of telephone calls form a normal distribution with a mean length of 8.0 min and a standard deviation of 2.5 min. The probability that a telephone call selected at random will last more than 15.5 min is most nearly

a.0.0013
b.0.0026
c.0.2600
d.0.9987

Answer 1

The answer is a.) 0.0013

Explanation:

Telephone calls of a normal distribution have a mean,
`\mu` = 8 min and a standard deviation,
`\sigma` = 2.5 min for the lengths of the call.

A randomly selected telephone call whose probability that the call will last more than 15.5 min will be given by

p ( x > 15.5)

The Z test statistic will Z =
`\frac{x} - \mu{\sigma} = ((15.5 - 8))/(2.5) = (7.5)/(2.5) = 3`.

Therefore we have to find p(Z > 3).

p(Z >3)
`\approx` 1 - p( Z < 3) = 1 - 0.9987 = 0.0013.

From the Z table we can find the value of p( Z < 3) to be 0.9987

The answer is a.) 0.0013