5. 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?

a. 0.9332

b. 0.0441 or 4.41%
c. 91.76

0.465

0.235
0.498
0.358

a. To find the probability that a randomly selected exam will have a score of at least 71, we need to calculate the area under the normal distribution curve to the left of 71.

First, we need to calculate the z-score for a score of 71 using the formula:

z = (x - μ) / σ

where x is the score, μ is the mean, and σ is the standard deviation.

z = (71 - 80) / 6
z ≈ -1.5

Next, we can use a standard normal distribution table or a calculator to find the probability associated with a z-score of -1.5. The probability is the area under the curve to the left of -1.5.

Looking up the z-score in the table or using a calculator, we find that the probability associated with a z-score of -1.5 is approximately 0.0668.

So, the probability that a randomly selected exam will have a score of at least 71 is approximately 0.0668, or 6.68%.

b. To find the percentage of exams that will have scores between 89 and 92, we need to calculate the area under the normal distribution curve between these two scores.

First, we need to calculate the z-scores for scores of 89 and 92 using the formula mentioned above:

z1 = (89 - 80) / 6
z1 ≈ 1.5

z2 = (92 - 80) / 6
z2 ≈ 2

Next, we can use a standard normal distribution table or a calculator to find the probabilities associated with these z-scores. The difference between these probabilities will give us the percentage of scores between 89 and 92.

Using the table or calculator, we find that the probability associated with a z-score of 1.5 is approximately 0.9332, and the probability associated with a z-score of 2 is approximately 0.9772.

So, the percentage of exams that will have scores between 89 and 92 is approximately (0.9772 - 0.9332) * 100 = 4.4%.

c. To find the lowest score eligible for an award, we need to find the z-score associated with the top 2.5% of test scores.

Using a standard normal distribution table or a calculator, we find that the z-score associated with the top 2.5% of the distribution is approximately 1.96.

Now, we can use the formula mentioned above to solve for the score:

z = (x - μ) / σ

1.96 = (x - 80) / 6

Solving for x, we get:

x - 80 = 1.96 * 6
x - 80 = 11.76
x ≈ 91.76

Therefore, the lowest score eligible for an award is approximately 91.76.

d. To determine the number of students who took the exam, we can use the information provided:

334 exams had scores of at least 89.

Using the normal distribution, we know that the percentage of scores at least as high as 89 is 1 - (probability associated with a z-score of 1.5) = 1 - 0.9332 = 0.0668.

Therefore, 0.0668 of the total number of students who took the exam scored at least 89.

Let's denote the total number of students as N.

0.0668 * N = 334

Solving for N, we get:

N = 334 / 0.0668 ≈ 5001

Therefore, approximately 5001 students took the exam.

a. To find the probability that a randomly selected exam will have a score of at least 71, we need to calculate the area under the normal distribution curve to the left of 71.

To do this, we can use a Z-table or a statistical calculator with the mean and standard deviation given.

The Z-score is calculated using the formula: Z = (X - μ) / σ

In this case, X = 71, μ = 80, and σ = 6.

Z = (71 - 80) / 6 = -1.5

Using a Z-table or calculator, we can find the area to the left of -1.5, which represents the percentage of scores below 71. Subtracting this value from 1 gives us the probability of a score of at least 71.

b. To find the percentage of exams that will have scores between 89 and 92, we need to calculate the area under the normal distribution curve between these two scores.

First, we calculate the Z-scores for both scores:

Z1 = (89 - 80) / 6 = 1.5
Z2 = (92 - 80) / 6 = 2.00

Using a Z-table or calculator, we can find the area to the left of each Z-score. Subtracting the smaller value from the larger value gives us the percentage of scores between 89 and 92.

c. To find the lowest score eligible for a merit award (top 2.5% of scores), we need to find the Z-score corresponding to this percentile.

Using a Z-table or calculator, we can find the Z-score that corresponds to the cumulative area of 0.975. This represents the area under the normal distribution curve to the left of the Z-score.

Once we have the Z-score, we can use the formula Z = (X - μ) / σ to solve for X, where X is the lowest score eligible for an award.

d. To find the number of students who took the exam, we need to use the information given - 334 exams with scores of at least 89.

Since we know the percentage of exams with scores of at least 89, we can calculate the total number of exams by setting up the equation:

(Number of exams with scores of at least 89) / (Total number of exams) = (Percentage of exams with scores of at least 89)

Solving for the total number of exams will give us the number of students who took the exam.