Question

The normaldistribution for women’s height in North America has a mean μ=62inches and a standard deviation ơ=2.75inches.The normal distribution for men’s height in North America has a mean μ=65inches and a standard deviation ơ=4 inches.Using the 68%, 95% and 99.7% Rule , determine in inches a). The women height interval fallingwithin 1standard deviation of the mean(Write the formula and show your work) b. The men height intervalfallingwithin 3standard deviation of the mean(Write the formula and show your work) c.(What percentage of the men heights falls between 57 inches and 73inches?(Write the formula and show your work)

Answer 1

Given that,

For Women : mean (μ) = 62 inches and standard deviation = 2.75 inches, Let X ~ N(62, 2.75)

For Men : mean (μ) = 65 inches and standard deviation = 4 inches, Let Y ~ N(65, 4)

According to (68-95-99.7) Rule,

i) Approximately 68% of the data fall within 1 standard deviations of the mean.

ii) Approximately 95% of the data fall within 2 standard deviations of the mean.

iii) Approximately 99.7% of the data fall within 3 standard deviations of the mean.

a)

Therefore, the women height interval falling within 1 standard deviations of the mean is from 59.25 inches to 64.75 inches.

b)

Therefore, men height Interval falling within 3 standard deviations of the mean is from 53 inches to 77 inches.

c)

Z = (57 - 65)/4 = -8/4 = -2 and Z = (73 - 65)/4 = 8/4 = 2

By rule ii) approximately 95% of the men heights falls between 57 inches and 73 inches.