Question

Consider the probability that exactly 89 out of 149 students will pass their college placement exams. Assume the probability that a given student will pass their college placement exam is 61%.Approximate the probability using the normal distribution

Answer 1

Answer: 0.0645

Step-by-step explanation:

From the information given,

probability of success, p = 61% = 61/100 = 0.61

q = 1 - p = 1 - 0.61 = 0.39

sample size, n = 149

Recall,

mean, μ = np = 149 x 0.61 = 90.89

Standard deviation, σ = √npq = √(90.89 x 0.39) = 5.95

Given that sample mean, x = 89, we would find P(x = 89)

By applying the continuity correction factor to the sample mean,

x = 89 would be 88.5 < x < 89.5

Thus,

P(x = 89) = P(88.5 < x < 89.5)

The formula for calculating the z score is

z = (x - μ)/σ

For x = 88.5,

z = (88.5 - 90.89)/5.95 = - 0.401

From the normal distribution table, the probability value corresponding to a z score of - 0.401 is 0.3446

For x = 89.5,

z = (89.5 - 90.89)/5.95 = - 0.23

From the normal distribution table, the probability value corresponding to a z score of - 0.23 is 0.4091

Thus,

P(88.5 < x < 89.5) = 0.4091 - 0.3446

P(88.5 < x < 89.5) = 0.0645