Question

a university administrator expects that 25% of students in a core course will receive an a. he looks at the grades assigned to 60 students. the probability that the proportion of students who receive an a is between 0.20 and 0.35 is

Answer 1

The correct option is b.

The probability that the proportion of students who receive an A is between 0.20 and 0.35 is approximately 0.5963.

To find the probability that the proportion of students who receive an A is between 0.20 and 0.35, we can use the normal distribution because we are dealing with a sample proportion. We'll need to calculate the z-scores for the lower and upper bounds and then find the probability in that range.

Step 1: Calculate the mean and standard deviation of the sample proportion.

Given:

• The expected proportion of students receiving an A is 0.25 (25%).
• The sample size is 60.

The mean of the sample proportion (μ) is equal to the expected proportion:

μ = 0.25

The standard deviation of the sample proportion (σ) is calculated using the formula:

σ = sqrt((p * (1 - p)) / n)

Where:

• p is the expected proportion (0.25).
• n is the sample size (60).

σ = sqrt((0.25 * (1 - 0.25)) / 60)

σ = sqrt((0.25 * 0.75) / 60)

σ = sqrt(0.01875 / 60)

σ ≈ 0.0860

Step 2: Calculate the z-scores for the lower and upper bounds.

To find the z-scores, we'll use the formula:

z = (X - μ) / σ

Where:

• X is the value we want to find the z-score for.
• μ is the mean (0.25).
• σ is the standard deviation (0.0860).

For the lower bound (X = 0.20):

z_lower = (0.20 - 0.25) / 0.0860 ≈ -0.5814

For the upper bound (X = 0.35):

z_upper = (0.35 - 0.25) / 0.0860 ≈ 1.1628

Step 3: Find the cumulative probabilities for the z-scores.

Now that we have the z-scores for the lower and upper bounds, we can use a standard normal distribution table or calculator to find the cumulative probabilities associated with these z-scores.

Using a standard normal distribution table or calculator:

• P(Z ≤ -0.5814) ≈ 0.2807
• P(Z ≤ 1.1628) ≈ 0.8770

Step 4: Calculate the probability that the proportion is between 0.20 and 0.35.

To find the probability that the proportion of students who receive an A is between 0.20 and 0.35, we subtract the cumulative probability at the lower bound from the cumulative probability at the upper bound:

P(0.20 ≤ X ≤ 0.35) = P(-0.5814 ≤ Z ≤ 1.1628) = P(Z ≤ 1.1628) - P(Z ≤ -0.5814) ≈ 0.8770 - 0.2807 ≈ 0.5963

So, the answer is approximately 0.5963.

The complete question is here:

A university administrator expects that 25% of students in a core course will receive an A. He looks at the grades assigned to 60 students.

The probability that the proportion of students who receive an A is between 0.20 and 0.35 is _________.

a. 0.1867

b. 0.5963

c. 0.8133

d. 0.9633

Answer 2

The probability that the proportion of students who receive an A is between 0.20 and 0.35 is 0.7766.

How to find the probability of the student

To solve this problem, use the normal approximation to the binomial distribution, since we have a large sample size (60 students) and can assume independence.

Given that the administrator expects 25% of students to receive an A, assume the probability of success (p) in each trial (student) is 0.25, and the probability of failure (q) is 1 - p = 0.75.

The proportion of students who receive an A can be approximated by a normal distribution with a mean of p and a standard deviation of
`\sqrt(pq/n)`, where n is the sample size.

Given, n = 60, p = 0.25, and q = 0.75.

To find the probability that the proportion of students who receive an A is between 0.20 and 0.35, calculate the z-scores for these proportions and use the standard normal distribution.

z₁ = (0.20 - 0.25) /
`\sqrt`((0.25 * 0.75) / 60)

z₂ = (0.35 - 0.25) /
`\sqrt`((0.25 * 0.75) / 60)

Using a standard normal distribution table, find the corresponding probabilities:

P(0.20 < p < 0.35) = P(z₁ < Z < z₂)

Now let's calculate the values:

Calculate the standard deviation:

σ =
`\sqrt`((0.25 * 0.75) / 60) ≈ 0.065

Calculate the z-scores:

z₁ = (0.20 - 0.25) / 0.065 ≈ -0.769

z₂ = (0.35 - 0.25) / 0.065 ≈ 1.538

Using the standard normal distribution table or a calculator:

P(0.20 < p < 0.35) = P(-0.769 < Z < 1.538) ≈ 0.7766

Therefore, the probability that the proportion of students who receive an A is between 0.20 and 0.35 is approximately 0.7766.

Complete question

A university administrator expects that 25% of students in a core course will receive an A. He looks at the grades assigned to 60 students.

The probability that the proportion of students who receive an A is between 0.20 and 0.35 is _________.

0.1867

0.7766

0.8133

0.9633