Answer:
The answer is a.) 0.0013
Step-by-step explanation:
Telephone calls of a normal distribution have a mean, [tex]\mu[/tex] = 8 min and a standard deviation, [tex]\sigma[/tex] = 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 = [tex]\frac{x} - \mu{\sigma} = \frac{(15.5 - 8)}{2.5} = \frac{7.5}{2.5} = 3[/tex].
Therefore we have to find p(Z > 3).
p(Z >3) [tex]\approx[/tex] 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
