Question

I need a calculation f a female that's 5.6 inches tall The average height for all males is 69.3 inches with a standard deviation of 2.8 inches.

For females, the average height is 64 inches and the standard deviation is 2.8. These are population values.

For this week’s discussion, you will calculate a z-score based on your own height and determine whether your score is within the 95% normal range or if it is out of that range and considered unusual. Measure your height as precisely as possible to a tenth of an inch. For example, 5’2 ¼” would be 62.3 inches tall. Calculate the z-score based on population values for males or females. Use the formula: z= (x-µ)/σ where x is your height in inches.

Calculate the normal range by creating the interval that is within 2 standard deviations of the population mean. Multiply the population standard deviation by 2 and then add/subtract from the population mean. In your discussion post, include the following, based on your calculations. Is your height within the normal range? Is this what you expected? Would your height be considered unusual? Why or why not.

Have you encountered any challenges based on your height? For example, someone who is shorter or taller may have a difficult time finding pants that are an appropriate length. How is the concept of normality used in your field? Example: a patient’s blood pressure is compared to a normal range of values and a financial planner may check the “average” return of a stock. How does knowing what is usual help people and corporations and government organizations plan? For example, how do airplane manufacturers use the normal height to design aircraft seating?

Answer 1

GIVEN THAT

According to the question we have the following data :

height for all males is 69.3 inches

standard deviation of 2.8 inches.

females, the average height is 64 inches

the standard deviation is 2.8

TO FIND :-    female that's 5.6 inches tall

NOW For female

the normal range for female is

now for finding for female that is 5.6 inch tall is

Z=5.6-64/2.8

= -20.86

as we know that 5.6inc <20.86

now finding standared deviation

Z=56-64/2.8

=-2.86

This is also not normal as 56 in is less than 2.86 standard deviation from the mean

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