Question

A population of women’s weights is approximately normally distributed with a mean of 140 lb and a standard deviation of 20 lb.

1. What percentage of women weigh between 120 and 160 lb?

2. If samples of size 16 are drawn from the population of women’s weights, what percentage of the sample means lie between 120 and 160 lb?

3. What is the probability that a sample mean from a sample size 16 lies above 145 lb?

Answer 1

Solution :

P(120 < x < 160) = P[(120 - 140)/ 20) < (x - ) /  < (160 - 140) / 20) ]

= P(-1 < z < 1)

= P(z < 1) - P(z < -1)

= 0.8413 - 0.1587

= 0.6826

Percentage = 68.26%

= / n = 20 / 16 = 5

= P[(120 - 140) / 5 < ( - ) / < (160 - 140) / 5)]

= P(-4 < Z < 4)

= P(Z < 4) - P(Z < -4)

= 1 - 0

= 1

Percentage = 100%

= / n = 20 / 16 = 5

P( > 145) = 1 - P( < 145)

= 1 - P[( - ) / < (145 - 140) / 5]

= 1 - P(z < 1)

= 1 - 0.8413

= 0.1587

Probability = 0.1587