Question

(1 point) Scores on a national 8th grade reading test are normally distributed with a mean of 130 and a standard deviation of 20. Find: (a) the probability that a single test score selected at random will be greater than 158: 0.080756659 (b) the probability that a random sample of 48 scores will have a mean greater than 132: 0.30724 (c) the probability that a random sample of 45 scores will have a mean greater than 132: (d) the probability that the mean of a sample of 23 scores will be either less than 124 or greater than 132:

Answer 1

Solution:- Given that mean = 130, sd = 20

a) P(X > 158) = P((X-mean)/sd > (158-130)/20)
= P(Z > 1.4)
= 0.0808

b) for n = 48

P(X > 132) = P((X-mean)/(sd/sqrt(n)) > (132-130)/(20/sqrt(48)))
= P(Z > 0.6928)
= 0.2451

c) for n = 45

p(X > 132) = P(Z > (132-130)/(20/sqrt(45)))
= P(Z > 0.6708)
= 0.2514

d) for n = 23

P(X < 124) = P(Z < (124-130)/(20/sqrt(45)))
= P(Z < -1.4387)
= 0.0749

P(X > 132) = P(Z > (132-130)/(20/sqrt(23)))
= P(Z > 0.4796)
= 0.3156

P(X < 124) + P(X > 132) = 0.0749 + 0.3156 = 0.3905