Question

Heights of adult women are distributed normally with a mean of 162 centimeters and a standard deviation of 7 centimeters.

Answer 1

Answer:

a) 0.75%

b) 60.9%

Explanation:

The heights of adult women are distributed normally with a mean(μ) = 162 centimetres and a standard deviation (σ) = 7 centimetres.

z score is a measure of the distance a raw score is from the mean in terms of standard deviation units. The z score is given by the equation:

$z=\frac{x-\mu}{\sigma}$.

a) The percentage of heights less than 145 centimetres. For x = 145 centimetres, the z score is:

$z=\frac{x-\mu}{\sigma} =\frac{145-162}{7}=-2.43$

P(x < 145) = P(z < -2.43) = 0.0075

The percentage of heights less than 145 centimetres is 0.75%

b) The percentage of heights between 165 centimetres and 180 centimetres

For x = 165 centimetres, the z score is:

$z=\frac{x-\mu}{\sigma} =\frac{160-162}{7}=-0.29$

For x = 180 centimetres, the z score is:

$z=\frac{x-\mu}{\sigma} =\frac{180-162}{7}=2.57$

P(160 < x < 180) = P(-0.29 < z < 2.57) = P(z < 2.57) - P(z < -0.29) = 0.9949 - 0.3859 = 0.609

The percentage of heights between 160 centimetres and 180 centimetres is 60.9%