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:
[tex]z=\frac{x-\mu}{\sigma} \\[/tex].
a) The percentage of heights less than 145 centimetres. For x = 145 centimetres, the z score is:
[tex]z=\frac{x-\mu}{\sigma} =\frac{145-162}{7}=-2.43[/tex]
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:
[tex]z=\frac{x-\mu}{\sigma} =\frac{160-162}{7}=-0.29[/tex]
For x = 180 centimetres, the z score is:
[tex]z=\frac{x-\mu}{\sigma} =\frac{180-162}{7}=2.57[/tex]
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%
