Question

Heights of 10 year old children, regardless of sex, closely follow a normal distribution with mean 54.1 inches and standard deviation 6.6 inches. Round answers to 4 decimal places. a) What is the probability that a randomly chosen 10 year old child is less than 49.2 inches? b) What is the probability that a randomly chosen 10 year old child is more than 62.4 inches? c) What proportion of 10 year old children are between 49.8 and 58.8 inches tall? d) 90% of all 10 year old child are less than inches.

Answer 1

Solution:- Given that mean 54.1 inches and standard deviation 6.6 inches

a) P(X < 49.2) = P((X-mean)/sd < (49.2-54.1)/6.6)
= P(Z < -0.7424)
= 1 − P(Z < 0.7424)
= 1 − 0.7704
= 0.2296

b) P(X > 62.4) = P((X-mean)/sd > (62.4-54.1)/6.6)
= P(Z > 1.2576)
= 1 − P(Z < 1.2576)
= 1 − 0.8962
= 0.1038

c) P(49.8 < X < 58.8) = P((49.8-54.1)/6.6 < Z < (58.8-54.1)/6.6))
= P(-0.6515 < Z < 0.7121)
= P(Z < 0.7121) − P(Z < −0.6515)
= 0.7611 - 0.2578
= 0.5033

d) for Z = 1.28

X = mean + Z*sd = 54.1 + 1.28*6.6 = 62.548