Question

In a class of 100 students, 25 students have hardcover and 75 students have paperback textbooks for the course. If you randomly choose 10 students in this class, fnd the probability that 2 of them have hardcover texts in the following ways: a. the exact probability b. approximate probability using a binomial distribution

Answer 1

To find the probability that 2 out of 10 randomly chosen students have hardcover textbooks, we can use the binomial probability formula. The exact probability is 0.275 and the approximate probability using a binomial distribution is 0.305.

Step-by-step explanation:

To determine the probability that exactly 2 out of 10 randomly chosen students have hardcover textbooks, we can use the binomial probability formula. The formula is P(X=k) = C(n,k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successes, p is the probability of success, and C(n,k) represents the combination of n items taken k at a time.

For this problem, n = 10 (number of students chosen), k = 2 (number of students with hardcover textbooks), and p = 0.25 (probability of a student having a hardcover textbook). Using these values, we can calculate the exact probability:

P(X=2) = C(10,2) * 0.25^2 * (1-0.25)^(10-2) = 45 * 0.25^2 * 0.75^8 = 0.2748, or approximately 0.275.

To find the approximate probability using a binomial distribution, we can use a calculator or a binomial distribution table. Using the same values as above, we can calculate the approximate probability to be 0.305.

Answer 2

h = prob of hardcover = .25

p = prob of paperback = .75

a.

Prob(2 hardcover of 10) = (10 choose 2) (.25)^2 (.75)^8

= ( 10(9)/2 ) (1/4^10) (1^2 3^8)

= 295245/1048576

≈ 28.15%

b.

I think I just did the binomial distribution. We can approximate it as a normal distribution. The mean and variance in this case are computed thus:

N=10, n=2

μ = Np = 2.5

σ² = Npq = 1.875

σ = 1.3693

P(n) = 1/(σ√2π)e^-(n-μ)/(2σ²)

P(2) = 1/(1.3693√2π)e^-(2 - 2.5)/(2 (1.875))

P(2) = 33.3%

Pretty good, not great.