Question

A doctor is using a growth chart for baby girls that indicates a 12-month-old baby girl has a mean weight of 21.1 pounds with a standard deviation of about 2.3 pounds. Assume that the weights are approximately normally distributed.

a) What is the probability that a randomly chosen 12-month-old baby girl weighs less than 19.58 pounds?

b) What is the probability that a randomly chosen 12-month-old baby girl weighs more than 22.85 pounds?

c) What proportion of 12-month-old baby girls weigh between 20.14 and 22.82 pounds?

Answer 1

Solution:- Given that mean = 21.1, sd = 2.3

a) P(X < 19.58) = P((X-mean)/sd < (19.58-21.1)/2.3)
= P(Z < -0.6609)
= 1 - P(Z < 0.6609)
= 1 - 0.7454
= 0.2546

b) P(X > 22.85) = P(Z > (22.85-21.1)/2.3)
= P(Z > 0.7609)
= 0.2236

c) P(20.14 < X < 22.82) = P((20.14-21.1)/2.3 < Z < (22.82-21.1)/2.3)
= P(-0.4174 < Z < 0.7478)
= P(Z < 0.7478) - P(Z < -0.4174)
= 0.7734 - 0.3372
= 0.4362