Question

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?

Answer 1

Answer:

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.