Answer the following. Show your work to get credit.

Exam scores are normally distributed with a mean of 81 and a standard deviation of 9.

a) What is the minimum score one must have to be in the top 4% of the students taking the exam?

b) What percentage of scores are below 90?

c) What percentage of scores are above 45?

d) If 30 people take this exam, how many would you expect to get above a 93%?

e) If you have a test score in the 45th percentile, what is its z-score?

f) What is the z-score and percentile for an exam score 1.2 standard deviations below the mean?

83%

a) To find the minimum score required to be in the top 4% of students, we need to find the z-score corresponding to the cutoff point.

The z-score formula is given by:
z = (X - μ) / σ

where z is the z-score, X is the raw score, μ is the mean, and σ is the standard deviation.

Since we want to find the minimum score, we need to find the z-score corresponding to the 4th percentile. The z-score associated with the 4th percentile is -1.75 (approximately) using a standard normal distribution table or a statistical calculator.

We can rearrange the formula to solve for X:
X = z * σ + μ
X = -1.75 * 9 + 81
X ≈ 64.75

Therefore, the minimum score required to be in the top 4% of students is approximately 64.75.

b) To find the percentage of scores below 90, we need to calculate the z-score for 90 using the same formula as before.

z = (90 - 81) / 9
z = 1

The z-score of 1 corresponds to approximately the 84th percentile. Since we want the percentage below 90, we subtract this from 100.

Percentage below 90 = 100 - 84
Percentage below 90 = 16%

Therefore, approximately 16% of scores are below 90.

c) To find the percentage of scores above 45, we again need to calculate the z-score for this specific score.

z = (45 - 81) / 9
z = -4

The z-score of -4 corresponds to an extremely low percentile close to 0. Since we want the percentage above 45, we subtract this from 100.

Percentage above 45 = 100 - 0
Percentage above 45 = 100%

Therefore, 100% of scores are above 45.

d) To find the number of people expected to get above a 93%, we need to calculate the z-score for this specific score.

z = (93 - 81) / 9
z ≈ 1.33

Looking this value up in a standard normal distribution table or using a calculator, we find that the percentage of scores above a z-score of 1.33 is approximately 90.8%.

Expected number of people = Percentage above 93% * Number of people taking the exam
Expected number of people = 0.908 * 30
Expected number of people ≈ 27.24

Therefore, we would expect approximately 27 people to get above a 93% score.

e) To find the z-score corresponding to a test score in the 45th percentile, we use the z-score formula and solve for z.

z = (X - μ) / σ

We know that the percentile corresponds to a z-score of approximately -0.125 (rounded). Rearranging the formula, we have:

-0.125 = (X - 81) / 9

Solving for X:

X - 81 = -0.125 * 9
X - 81 = -1.125
X ≈ 79.875

Therefore, a test score in the 45th percentile would be approximately 79.875, which we can round to 80 for practical purposes.

To calculate the z-score for this test score, we use the formula:

z = (X - μ) / σ
z = (80 - 81) / 9
z ≈ -0.111

Therefore, the z-score for a test score in the 45th percentile is approximately -0.111.

f) To find the z-score and percentile for an exam score 1.2 standard deviations below the mean, we can use the formula:

z = (X - μ) / σ

Substituting the given values, we have:

z = (X - 81) / 9 = -1.2

Solving for X:

(X - 81) / 9 = -1.2
X - 81 = -1.2 * 9
X - 81 = -10.8
X ≈ 70.2

Therefore, the score 1.2 standard deviations below the mean is approximately 70.2.

To calculate the percentile, we can use a standard normal distribution table or a calculator. The z-score of -1.2 corresponds to approximately the 11.86th percentile.

Therefore, the z-score for an exam score 1.2 standard deviations below the mean is approximately -1.2, and the percentile is approximately 11.86%.