Suppose the length 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 telephone call selected a random will last more than 15.5 min is most nearly

(A) 0.0013
(B) 0.0026
(C) 0.2600
(D) 0.9987

(A) 0.0013

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The probability that telephone call selected a random will last more than 15.5 min is most nearly

This is 1 subtracted by the pvalue of Z when X = 15.5.

So

has a pvalue of 0.9987

1 - 0.9987 = 0.0013

So the correct answer is:

(A) 0.0013

correct answer:

b.(19,-2)

correct choice is b

step-by-step explanation:

when you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. for the function the average rate of change from point to point can be calculated as

to find the average rate of change from day 2 to day 10, note that at day 2, 8 houses were sold, so a(2,8) and at day 10, 2 houses were sold, so b(10,2). thus,