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

The answer is a.) 0.0013

Step-by-step explanation:

Telephone calls of a normal distribution have  a mean, = 8 min and a standard deviation, = 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   = .

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

p(Z >3)  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

$30.32

step-by-step explanation:

65.00*.12=2.80

65.00-2.80=57.20

57.20*1.06=60.632

60.632/2=30.316

ans789

step-by-step explanation: