The chance that a college student has a job offer before graduation is 70%. If I sample 200 graduates… (use continuity correction)

a) what is the chance more than 130 of them have a job offer?

b) what is the chance less than 133 of them have a job offer?

c) what is the chance at least 150 of them have a job offer?

d) what is the chance at most 144 of them have a job offer?

To solve these questions, we can use the normal distribution with a continuity correction. Let's start by calculating the mean and standard deviation:

Given:
Probability of having a job offer before graduation = 0.70
Sample size (n) = 200

a) To find the chance that more than 130 of them have a job offer, we need to calculate the z-score and use the standard normal distribution table.

Mean (μ) = n * p = 200 * 0.70 = 140
Standard Deviation (σ) = sqrt(n * p * (1 - p)) = sqrt(200 * 0.70 * (1 - 0.70)) = sqrt(42)

Continuity Correction:
To use the continuity correction, we need to add/subtract 0.5 from the boundaries.

Z-score = (130 - 0.5 - 140) / sqrt(42) = -1.43

Using the standard normal distribution table or a calculator, we find that the probability is 1 - P(Z < -1.43) = 1 - 0.0764 = 0.9236. Therefore, the chance that more than 130 of them have a job offer is approximately 92.36%.

b) To find the chance that less than 133 of them have a job offer, we also need to calculate the z-score.

Z-score = (133 + 0.5 - 140) / sqrt(42) = -0.82

Using the standard normal distribution table or a calculator, we find that the probability is P(Z < -0.82) = 0.2054. Therefore, the chance that less than 133 of them have a job offer is approximately 20.54%.

c) To find the chance that at least 150 of them have a job offer, we can calculate the z-score and its complement.

Z-score = (150 - 0.5 - 140) / sqrt(42) = 1.37

Using the standard normal distribution table or a calculator, we find that the probability is 1 - P(Z < 1.37) = 1 - 0.9131 = 0.0869. Therefore, the chance that at least 150 of them have a job offer is approximately 8.69%.

d) To find the chance that at most 144 of them have a job offer, we can again calculate the z-score and its complement.

Z-score = (144 + 0.5 - 140) / sqrt(42) = 0.86

Using the standard normal distribution table or a calculator, we find that the probability is P(Z < 0.86) = 0.8051. Therefore, the chance that at most 144 of them have a job offer is approximately 80.51%.

To solve these probability questions, we will use the normal distribution and apply continuity correction since we are dealing with a discrete probability event. Let's go through each question step by step:

a) To find the probability that more than 130 graduates have a job offer out of a sample of 200, we need to calculate the area under the normal curve to the right of 130.

Step 1: Convert the problem to a standard normal distribution by using the formula:
z = (x - μ) / σ

In this case, x = 130, μ = np (where n is the sample size and p is the probability of success), and σ = √(npq) (where q = 1 - p).
Since we are given that the chance of job offers is 70% (or 0.7) for college students, we have μ = 200 * 0.7 = 140 and σ = √(200 * 0.7 * 0.3) = 8.16.

Substituting the values into the formula, we get:
z = (130 - 140) / 8.16 = -1.22

Step 2: Use the standard normal distribution table or a calculator to find the area to the right of z = -1.22. The area represents the probability of more than 130 graduates having a job offer.

Using the standard normal distribution table or a calculator, we find that the area to the right of z = -1.22 is approximately 0.8888.

Therefore, the probability that more than 130 graduates have a job offer is approximately 0.8888.

b) To find the probability that less than 133 graduates have a job offer out of a sample of 200, we need to calculate the area under the normal curve to the left of 133.

Step 1: Convert the problem to a standard normal distribution by using the formula:
z = (x - μ) / σ

In this case, x = 133, μ = np, and σ = √(npq). Using the same values for μ and σ as in part a), we have:
z = (133 - 140) / 8.16 = -0.857

Step 2: Use the standard normal distribution table or a calculator to find the area to the left of z = -0.857. The area represents the probability of less than 133 graduates having a job offer.

Using the standard normal distribution table or a calculator, we find that the area to the left of z = -0.857 is approximately 0.1957.

Therefore, the probability that less than 133 graduates have a job offer is approximately 0.1957.

c) To find the probability that at least 150 graduates have a job offer out of a sample of 200, we need to calculate the area under the normal curve to the right of 150.

Step 1: Convert the problem to a standard normal distribution by using the formula:
z = (x - μ) / σ

In this case, x = 150, μ = np, and σ = √(npq). Again, using the same values for μ and σ as in parts a) and b), we have:
z = (150 - 140) / 8.16 = 1.23

Step 2: Use the standard normal distribution table or a calculator to find the area to the right of z = 1.23. The area represents the probability of at least 150 graduates having a job offer.

Using the standard normal distribution table or a calculator, we find that the area to the right of z = 1.23 is approximately 0.1093.

However, since we want the probability of "at least" 150 graduates having a job offer, we need to subtract this value from 1:
1 - 0.1093 ≈ 0.8907

Therefore, the probability that at least 150 graduates have a job offer is approximately 0.8907.

d) To find the probability that at most 144 graduates have a job offer out of a sample of 200, we need to calculate the area under the normal curve to the left of 144.

Step 1: Convert the problem to a standard normal distribution by using the formula:
z = (x - μ) / σ

In this case, x = 144, μ = np, and σ = √(npq). Once again, using the same values for μ and σ as in parts a), b), and c), we have:
z = (144 - 140) / 8.16 = 0.49

Step 2: Use the standard normal distribution table or a calculator to find the area to the left of z = 0.49. The area represents the probability of at most 144 graduates having a job offer.

Using the standard normal distribution table or a calculator, we find that the area to the left of z = 0.49 is approximately 0.6844.

Therefore, the probability that at most 144 graduates have a job offer is approximately 0.6844.