Mathematics, 13.10.2019 23:30 weirdojuwin

Scores on a recent national statistics exam were normally distributed with a mean of 80 and a standard deviation of 6. a. what is the probability that a randomly selected exam will have a score of at least 71? b. what percentage of exams will have scores between 89 and 92? c. if the top 2.5% of test scores receive merit awards, what is the lowest score eligible for an award? d. if there were 334 exams with scores of at least 89, how many students took the exam? e. what is the median of the population of the recent national statistics exam?

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Answer from: morgannwaldrupp

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 $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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, $\mu = 80$ .Standard deviation of 6, thus, $\sigma = 6$ .

Item a:

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{71 - 80}{6}$

Z = -1.5

Z = -1.5 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{92 - 80}{6}$

Z = 2

Z = 2 has a p-value of 0.9772.

X = 89

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{89 - 80}{6}$

Z = 1.5

Z = 1.5 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.

$Z = \frac{X - \mu}{\sigma}$

$1.96 = \frac{X - 80}{6}$

X - 80 = 6(1.96)

X = 91.76

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.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{89 - 80}{6}$

Z = 1.5

Z = 1.5 has a p-value of 0.9332.

1 - 0.9332 = 0.0668.

Thus, 0.0668 of n is 334, thus:

0.0668n = 334

$n = \frac{334}{0.0668}$

n = 5000

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

Answer from: jermainedwards

0.0668, 0.0440, 92, 5000, 80

Step-by-step explanation:

a. First calculate the z-score:

z = (x - μ) / σ

z = (71 - 80) / 6

z = -1.5

Now we use a calculator or z-score table to find P(z<-1.5).

P(z<-1.5) = 0.0668

b. Again, we calculate the z-scores:

z = (x - μ) / σ

z = (89 - 80) / 6

z = 1.5

z = (x - μ) / σ

z = (92 - 80) / 6

z = 2

Using a calculator or z-score table, we want to find P(1.5<z<2). Which is equal to P(z<2) - P(z<1.5).

P(z<2) - P(z<1.5)

0.9772 - 0.9332

0.0440

c. This time we want to go backwards. We need to find the z score such that P(z>?) = 0.0250. Or P(z<?) = 1 - 0.0250 = 0.9750. According to the table, z = 1.96.

Now finding the x value that gives us that z-score:

z = (x - μ) / σ

1.96 = (x - 80) / 6

11.76 = x - 80

x = 91.76

Since scores have to be whole numbers, we round up to 92.

d. From part b, we know the z-score for this is 1.5, and that P(z<1.5) = 0.9332. That means that P(z>1.5) = 1 - 0.9332 = 0.0668.

So if 334 scores make up 6.68% of all scores, then the number of students who took the exam is:

0.0668 N = 334

N = 5000

e. For a normal distribution, the median is also the mean. 80.

Answer from: Quest