Demonstrate the use of the normal distribution, the standard normal distribution

Question

Demonstrate the use of the normal distribution, the standard normal distribution, and the central limit theorem for calculating areas under the normal curve and exploring these concepts in real life applications.

Frank has only had a brief introduction to statistics when he was in high school 12 years ago, and that did not cover inferential statistics. He is not confident in his ability to answer some of the problems posed in the course.

Concept Being Studied

Finding areas and probabilities by using the standard normal distribution and the Z-Table.

As Frank's help, you need to provide Frank with guidance and instruction on a worksheet he has partially filled out. Your job is to help him understand and comprehend the material. You should not simply be providing him with an answer as this will not help when it comes time to take the test. Instead, you will be providing a step-by-step breakdown of the problems including an explanation on why you did each step and using proper terminology.

What to Submit

To complete this assignment, you must first download the worksheet, and then complete it by including the following items on the worksheet:

Incorrect Answers

Correct any wrong answers. You must also explain the error performed in the problem in your own words.

Partially Finished Work

Complete any partially completed work. Make sure to provide step-by-step instructions including explanations.

Blank Questions

Show how to complete any blank questions by providing step-by-step instructions including explanations.

Your step-by-step breakdown of the problems, including explanations, should be present within the answers provided on the document.

Deliverable 02 – Worksheet

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.

<p> Assume that a randomly selected subject is given a bone density test. Those tests are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that the bone density score for this subject is between -1,93 and 2.37

Student’s answer: We first need to find the areas under the Z table for these values.

For -1.93 the area from the left is 0.268, and for 2.37 the area is .9911

Continue the solution:

Finish the problem giving step-by-step instructions and explanations.

<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 63.8 in. and a standard deviation of 2.6 in, find the percentage of women that meet the height requirement.

Answer and Explanation:

Enter your step-by-step answer and explanations here.

<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 z = (x-)/

z = (69.4 - 66)/11.3

z = .301 (rounded)

            Corrections:

Enter your corrections and explanations here.

<b> If we look up the z value of -1.645 in the Z table what is the cumulative area from the left under the curve, and what is the area on the right of that value?

Enter your step-by-step answer and explanations here.

<i> What is the Z value for an area under the standard normal distribution of 0.8980 from the right?

Student’s answer: looking at the positive Z table, we find an area of .8980 and we see that the Z value is 1.27

Corrections:

Enter your corrections and explanations here.

<b> 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 1.0 in. Assume that a manhole cover is constructed with a diameter of 22 in. What percentage of men will fit into a manhole with this diameter?

Answer and Explanation:

Enter your step-by-step answer and explanations here.

Explanation / Answer

A) Mean =0 ; Std = 1

P(-1.93>X<2.37)

Z1=(x-mean)/std = (-1.93-0)/1 = -1.93

Z2= (2.37-0)/1 2.37

P(Z1<-1.93) = 0.5-0.47320= 0.0268 (using left hand distribution table)

P(Z2<2.37)= 0.99086

P(-1.93> X<2.37) = 0.99086-0.0268 =0.9640

B)US Air force Pilot heights mean =64 and 77

If women’s height of xi= 63.8, std = 2.6

Women’s height Mean = 64

Z=(xi-Mean)/std = (63.8-64)/2.6 = -0.07692

P(Z<-0.07692) = 0.5-0.02790= 0.4721

P(Z<63.8) = 0.4721

C) Mean =69.4

Std = 11.3

Xi =66

Z = (Xi-Mean) / std = (66-69.4)/11.3 = -0.30088

Where negative representation refers Left hand of Standard normal distribution.

D)Z = -1.645

P(Z<1.645) = 0.95

Here, Probability of Area of Normal distribution = 1

P(Z< -1.645)= 1-0.95 = 0.05 (Left of the Curve)

P(Z>-1.645) = 1-P(Z<-1.645) = 1-0.05 = 0.95 (Right of the Curve)

E)Standard normal distribution value = 0.8980

From Right of standard distribution table covers 50% area.

Z(0.8980)= Z(0.8+0.0980) = 0.31327

Z(X>0.8980) mea Right side of the data = 0.5-0.31327 = 0.18673

F) Min Xi =22, Max= 60

Mean = 18.2, std = 1

Z=(xi-Mean)/std

Z = (22-18.2)/1 = 3.8

Where as P(X>22) = 1-

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