18. A normal population has a mean of 80.0 and a standard deviation of 14.0. a. Compute the probability of a value between 75.0 and 90.0. b. Compute the probability of a value 75.0 or less. c. Compute the probability of a value between 55.0 and 70.0. Lind, Douglas. Basic Statistics for Business and Economics (p. 213). McGraw-Hill Education. Kindle Edition.

a) 40.17% probability of a value between 75.0 and 90.0.

b) 35.94% probability of a value 75.0 or less.

c) 20.22% probability of a value between 55.0 and 70.0.

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:

a. Compute the probability of a value between 75.0 and 90.0.

This is the pvalue of Z when X = 90 subtracted by the pvalue of Z when X = 75.

X = 90

has a pvalue of 0.7611

X = 75

has a pvalue of 0.3594

0.7611 - 0.3594 = 0.4017

40.17% probability of a value between 75.0 and 90.0.

b. Compute the probability of a value 75.0 or less.

This is the pvalue of Z when X = 75. So

has a pvalue of 0.3594

35.94% probability of a value 75.0 or less.

c. Compute the probability of a value between 55.0 and 70.0.

This is the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 55.

X = 70

has a pvalue of 0.2389

X = 55

has a pvalue of 0.0367

0.2389 - 0.0367 = 0.2022

20.22% probability of a value between 55.0 and 70.0.

the answer is 8

step-by-step explanation:

3.14 x 8x8