Deliverable 2 - Tutoring on the Normal Distribution

Instructions: The following worksheet is shown to you by a student who is asking for help. Your job is to help the student walk through the problems by showing the student how to solve each problem in detail. You are expected to explain all of the steps in your own words.

Key:

· <i> - This problem is an incorrect. Your job is to find the errors, correct the errors, and explain what they did wrong.

· <p> - This problem is partially finished. You must complete the problem by showing all steps while explaining yourself.

· <b> - This problem is blank. You must start from scratch and explain how you will approach the problem, how you solve it, and explain why you took each step.

1) <p> Assume that a randomly selected subject is given a bone density test. Those tests follow a standard normal distribution. Find the probability that the bone density score for this subject is between -1.53 and 1.98

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:

The equation that we will use for this problem will be P (-1.53 < z < 1.98). The answer listed for the normal distribution of -1.53 being 0.0630 and 1.95 being 0.9761 is correct.

To finish the problem, we will need to solve z; we need to find the probability that the bone density sore is between -1.53 and 1.98 we will subtract the larger probability from the smaller probability.

The excel formula would be: =norm.dist(1.98,0,1,true)-norm.dist(-1.53,0,1,true)= 0.9131 converted to 91.31%.

Finally, the completed problem will provide us with the probability that there would be a bone density score between -1.53 and 1.98 of 91.31% which is a high probability.

Mean=0

STD= 1

2) <b> 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:

Our values are: mean = 65 standard deviation = 3.5 height between 64 in. and 77 in

For this question, we need to find the percentage of women that meet the height requirement for the U.S. Airforce that are between 64inches and 77inches. We will write the problem as P (64 < z < 77) as we are finding the probability between two number values. The first step will be to find the normal distribution for both 64 and 77.

The excel function that we will utilize is; =NORM.DIST

So, for the height of 64, the function would be =NORM.DIST(64, 65, 3.5, true) because the mean is 65 and the standard deviation is 3.5 in this problem, and you end up with an answer of 0.3875. To solve for 77, you will enter the formula into excel as =NORM.DIST(77, 65, 3.5, true), which gives you an answer of 0.9997. When you subtract these two numbers 0.9997-0.387, you get .6122.

Since we are using excel to work the problem, we will follow the below formula to get the same answer.

To complete the problem and solve for the percentage of women that meet the height requirement between 64inches and 77inches, you will have to subtract the larger probability from the smaller probability. So, the formula for excel will be =NORM.DIST(77, 65, 3.5, true)- NORM.DIST(64, 65, 3.5, true). The answer will be 0.6121.

Since we want to convert the answer into a percentage, we will move the decimal point over to the right two spaces which will change the answer to 61.21.

We finally have an answer which tells us that the percentage of women who meet the height requirement between 64 and 77 inches for the U.S. Airforce is 61.21%.

3) <i> 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:

Let

Corrections:

The standard deviation is 11.3 The mean is 69.4 And x=66

The basic equation given is correct

In this equation μ = mean and σ = standard deviation.

The error in this problem is that when the numbers were plugged into the equation, they were entered in the wrong order. The answer provided has the original value 66 and being subtracted from the mean of 69.4. The correct order should be the mean 69.4 being subtracted from the original value of 66.

This how it should look: Z=

The formula to enter in excel will be = (66-69.4)/11.3 Z= -0.300885

For Z scores you round to two decimals, and therefore it would change to -0.30

The answer for the z-score for a woman having a pulse rate of 66 beats per minute as Z-Score = -0.30.

4) <b> 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:

Since we are using standard normal distribution, the mean will be 0, and the standard deviation will be 1.

To find the z value to the left, you would solve P (z < x). Which means the probability that z is less than x. Next, we will need to solve for the area on the right; you would solve P (z > x), which means z is greater x.

The first part of the equation we will look at is for the left side. To find the cumulative area, we will use P (z < -0.875). The function we would need to enter in excel would be =NORM.DIST(-0.875, 0, 1, true), where the mean = 0 and the standard deviation = 1. This gives us an answer of 0.1908.

To solve for the value on the right, we will use the same excel function except we will add 1- to the formula. So, the formula we will use is =NORM.DIST(1- -0.875, 0, 1, true) which will give us the answer of 0.9696. In this formula since we are finding the value greater than a number, we must take away the value by using the 1- in the excel formula.

Our answer to the question is that the probability that z is less than (to the left) of -0.875 is 19.08%, and the probability that z is greater than (to the right) of -0.875 is 96.96%.

The

5) <i> 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:

For this problem, we need to find the Z-score, and we know that the probability is 0.6573.

The Mean is 0, and the standard deviation is 1. The excel function that we will use for this problem is NORM.INV

First, we need to look at the area on the right, so we know that we are looking for is the probability that some standard normal variable will be greater than our z-score.

This is problem is written as: P (Z > z) =0.6573.

Since we are working with the probability that some value will be larger than our z-score, we would have to subtract one from our probability in our formula NORM.INV (1-.0.6573,0,1) which gives -0.41 which equals= Z.

What this answer means is that the probability that some standard normal variable will be greater than 0.6573 is -0.41

So: P (Z>-0.41) =0.6573

6) <p> 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:

For this problem, we need to find the percentage of men that will fit into a manhole that has a diameter of 22.5 inches. To find the percentage for this problem, we first need to find the probability of men having shoulders with a width less than or equal to 22.5 inches. The equation we will use for this problem is P(z<22.5).

The equation used for normal distribution in excel is =NORM.DIST.

For this problem, X= 22.5 Mean is 18.2, and the standard deviation is 2.09

We will enter the problem into excel as:

=NORM.DIST(22.5, 18.2, 2.09, TRUE)

The answer we get will be 0.9655 which equals 96.55%

Therefore, the percentage of men who would fit into a 22.5in diameter manhole is 96.55%