Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Question 3 Suppose men\'s height is normally distributed with mean 69 in. and st

ID: 3057362 • Letter: Q

Question

Question 3

Suppose men's height is normally distributed with mean 69 in. and standard deviation 3 in. Women's height is normally distributed with mean 64 in. and standard deviation 2 in.

John is 73 in. tall and Mary is 62 in.

Question 3 options:

A. _____

If JANE is considered with a height of 67 in., who is MORE EXTREME relative to their population John or Jane?

B. _____

Who is MORE EXTREME relative to their gender populations John or Mary?

C. _____

Who was closest to their population's 50th percentile John or Mary?

What is John's z-score?

E. _____

What is Mary's z-score?

Note: Below choices may NOT all be used as matches.

Mary

-1.00

1.33

1.00

No answer is correct.

John and Mary have the same position in their populations.

Jane

John

.

A. _____

If JANE is considered with a height of 67 in., who is MORE EXTREME relative to their population John or Jane?

B. _____

Who is MORE EXTREME relative to their gender populations John or Mary?

C. _____

Who was closest to their population's 50th percentile John or Mary?

D. _____

What is John's z-score?

E. _____

What is Mary's z-score?

1.

Note: Below choices may NOT all be used as matches.

2.

Mary

3.

-1.00

4.

1.33

5.

1.00

6.

No answer is correct.

7.

John and Mary have the same position in their populations.

8.

Jane

9.

John

Explanation / Answer

I think John is male and Mary and Jale are female..

we find z-score for them and z=(x-mean)/sd

for John, z=(73-69)/3=1.33

for mary, z=(62-64)/2=-1

for Jane, z=(67-64)/2=1.5

(A)Jane, as Jane z-score is more than John

(B)John, as absolute z-score is more for John

(C) mary, because mary z score= -1 is near to 0 than John

(D)John, z=(73-69)/3=1.33

(E)-1

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote