I have answered the following questions but need help in inputting these into an excel sheet:

Student's answer: We first need to find the probability for each of these z-scores using Excel.

For -1.53 the probability from the left is 0.0630, and for 1.98 the probability from the left is 0.9761.

Continue the solution:

Once the probability for both variables, you then subtract the larger variable from the smaller one.

0.9746148-0.063008= 0.91314

The z score only needs 2 decimal places. Z = 0.91

2) The U.S. Airforce requires that pilots have a height between 64 in. and 77 in. If women's heights are normally distributed with a mean of 65 in. and a standard deviation of 3.5 in, find the percentage of women that meet the height requirement.

Answer and Explanation:

We must first find the probability for each variable in the excel sheet. There s a height requirement and we need to find out how many women meet that requirement.

0.387548 is left of 64

0999697 is left of 77

We now subtract the smaller variable from the larger variable

0.999697-0.387548 = 0.612148

We now need to find the percentage by moving the decimal 4 places to the right.

61.21% of the women meet the height requirement

3) Women's pulse rates are normally distributed with a mean of 69.4 beats per minute and a standard deviation of 11.3 beats per minute. What is the z-score for a woman having a pulse rate of 66 beats per minute?

Student's answer:

Corrections:

The variable and mean have been placed in the wrong spots, the correct answer is:

Z - (66-69.4)/11.3 - 0.30088

Z = -0.30

4) What is the cumulative area from the left under the curve for a z-score of -0.875? What is the area on the right of that z-score?

Answer and Explanation:

We must first find the cumulative area from the left.

P(z,-0.875) Area to the left is = 19.08%

We must also find the cumulative area from the right.

1-p(z < -0.875) Area to the right is = 80.92%

5) If the area under the standard normal distribution curve is 0.6573 from the right, what is the corresponding z-score?

Student's answer: We plug in "=NORM.INV(0.6573, 0, 1)" into Excel and get a z-score of 0.41.

Corrections:

The correct probability formula should be = NORM.S.INV

We now must find the left.

We take NOR.S.INV(0.6573) and put into excel sheet to find the z score, which is 0.405105. We then round to decimal places. The answer is 0.41

6)

Manhole covers must be a minimum of 22 in. in diameter, but can be as much as 60 in. Men have shoulder widths that are normally distributed with a mean of 18.2 and a standard deviation of 2.09 in. Assume that a manhole cover is constructed with a diameter of 22.5 in. What percentage of men will fit into a manhole with this diameter?

Student's answer: We need to find the probability that men will fit into the manhole. The first step is to find the probability that the men's shoulder is less than 22.5 inches.

Continue the solution:

We must now see if the probability of the men's shoulder is less than 22.5 in by using NORM.DIST(x,mean,standard,deviation,true) formula. This equals 0.980177.

98.02% of men will fit into 22.5 inch manhole.

To find out how many men would not fit in the 22.5 inch manhole we must now take 1 from 0.980177. This equals 0.019823. 1.98% of men can not fit in the 22.5 manhole and would require a larger hole.

PART 2: MY QUESTION

Please help! I do not know how to put my answers/calculations into an excel sheet. Can you please help me to put them into an excel sheet so I can submit my assignment ? I do not really understand how how or what it is asking for when it says put into an excel sheet.

Thank you