Question

The scores of students on the ACT (American College Testing)college entrance examination in a recent year had the normal distribution with meanμ= 18 and standard deviationσ= 6. 100 students are randomly selected from all who took the test

a.What is the probability that the mean score for the 100 students is between 17and 19 (including 17 and 19)?

b.A student is eligible for an honor program if his/her score is higher than 25.Find an approximation to the probability that at least 15 students of the 100 students are eligible for the honor program.

c.If the sample size is 4 (rather than 100), what is the probability that more than50% (not include 50%) students are eligible for the honor program?

Answer 1

Answer:

a)

P(17 < < 19)

= P(-1.67 < Z < 1.67)

= P(Z < 1.67) - P(Z < -1.67)

= 0.9525 - 0.0475

= 0.9050

b)

P(X > 25)

= P(Z > 1.17)

= 1 - P(Z < 1.17)

= 1 - 0.8790

= 0.1210

n = 100

p = 0.121

= np = 100 * 0.121 = 12.1

= sqrt(np(1 - p))

= sqrt(100 * 0.121 * (1 - 0.121))

= 3.2613

P(X > 15)

= P((X - )/ > (14.5 - )/)

= P(Z > (14.5 - 12.1)/3.2613)

= P(Z > 0.74)

= 1 - P(Z < 0.74)

= 1 - 0.7704

= 0.2296

c)

n = 4

p = 0.121

It is a binomial distribution.

P(X = x) = nCx * px * (1 - p)n - x

P(X > 2) = P(X = 3) + P(X = 4)

= 4C3 * (0.121)^3 * (0.879)^1 + 4C4 * (0.121)^4 * (0.879)^0

= 0.0064

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