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. Scenario 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. As Frank's tutor, 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 word document, 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 word document provided. You must also include an Excel workbook which shows all your calculations performed.

Answer 1

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-