Which number goes in the blank to make the statement true? 3.098 < it a) 3.089 b) 3.908 c) 2.099 or d) 2.998

B. because it's 3 and 9 TENTHS not 3 and 9 HUNDRETHS, place value

Using the normal distribution, it is found that:

a) 0.9332 = 93.32% probability that a randomly selected exam will have a score of at least 71.

b) 4.4% of exams will have scores between 89 and 92.

c) The lowest score eligible for an award is of 91.76.

d) 5000 students took the exam.

e) The median of the population is of 80.

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

It 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, which is the percentile of X.

In this problem:

Mean of 80, thus, .Standard deviation of 6, thus, .

Item a:

This probability is 1 subtracted by the p-value of Z when X = 71, thus:

has a p-value of 0.0668.

1 - 0.0668 = 0.9332.

0.9332 = 93.32% probability that a randomly selected exam will have a score of at least 71.

Item b:

The proportion is the p-value of Z when X = 92 subtracted by the p-value of Z when X = 89, thus:

X = 92

has a p-value of 0.9772.

X = 89

has a p-value of 0.9332.

0.9772 - 0.9332 = 0.044

0.044 x 100% = 4.4%

4.4% of exams will have scores between 89 and 92.

Item c:

The lowest score eligible is the 100 - 2.5 = 97.5th percentile, which is X when Z has a p-value of 0.975, so X when Z = 1.96.

The lowest score eligible for an award is of 91.76.

Item d:

First, we find the proportion of scores above 89, which is 1 subtracted by the p-value of Z when X = 89.

has a p-value of 0.9332.

1 - 0.9332 = 0.0668.

Thus, 0.0668 of n is 334, thus:

5000 students took the exam.

Item e:

In the normal distribution, the median is the same as the mean, thus the median of the population is of 80.

A similar problem is given at link