Do this one by hand. Suppose we measured the height of 10,000 men and found that

Question

Do this one by hand. Suppose we measured the height of 10,000 men and found that the data were normally distributed with a mean of 70.0 inches and a standard deviation of 4.0 inches. Answer the following questions and show your work:

What proportion of men can be expected to have heights less than: 66, 70, 72, 75 inches?

What proportion of men can be expected to have heights greater than: 64, 66, 73, 78 inches?

What proportion of men can be expected to have heights between: 65 and 75 inches, 71 and 72 inches?

How many men (of the 10,000) can be expected to have heights between: 65 and 71 inches, 72 and 75 inches?

What height corresponds to the 43rd percentile? What height corresponds to the 55th percentile? What height corresponds to the 99th percentile? (If necessary, round the numbers, but keep at least two decimal places.)

I need 1E, 2 and 3!!

Do this one by hand. Suppose we measured the height of 10,000 men and found that the data were normally with a mean of 70.0 inches and a standard deviation of 4.0 inches. Answer the following questions and show your work: A. What proportion of men can be expected to have heights less than: 66, 70, 72, 75 inches? B. What proportion of men can be expected to have heights greater than: 64, 66, 73, 78 inches? C. What proportion of men can be expected to have heights between: 65 and 75 inches. 71 and 72 inches? D. How many men (of the 10,000) can be expected to have heights between: 65 and 71 inches, 72 and 75 inches? E. What height corresponds to the 43rd percentile? What height corresponds to the 55, h percentile? What height corresponds the 99th percentile? (If necessary, round the numbers, but keep at least two decimal places.) Suppose for the same 10,000 men, we also measured their weight. Suppose the data were again normally distributed. The average weight is 175.0 lbs., and the standard deviation is 25.0 lbs. Suppose the correlation between height and weight is r = +.61. Answer the following and show your work: A. What is the slope of the best-fit line to predict height from weight? What is the intercept of the line? (Make height the Y variable and weight the X variable). B. Write a sentence or two that says what the equation of the line tells you about the relation between the variables: that is, when weight increases by one pound, how much does the predicted value of height increase? C. What is your best guess for the height of a man who weighs 150 lbs.? For the height of a man who weighs 175 lbs.?

Explanation / Answer

1E) here we use standard normal variate z=(x-mean)/sd=(x-70)/4

we want to find x, such that P(X<x)=0.43

43rd percentile

for this we find z value such that P(Z<z)=0.43 and z=-0.1764 ( use ms-excel command==NORMSINV(0.43))

so x=70-4*0.1764=0.6929

55 percentile

z=0.1257 and x=70+4*0.1257=70.50

99th percentile

z=2.3264 and x=70+4*2.3264=79.31

(2) mean(height)=70, sigma(height)=4

mean(weight)=175, sigma(weight)=25

corr(height,weight)=r=0.61

(a)slope=r*sigma(height)/sigma(weight)=0.61*4/25=0.0976

intercept=mean(height) -slope*mean(weight)=70-0.0976*175 =52.92

height=52.92+0.0976*weight

(b) slope is per unit change in dependent variable (height) when unit change is done in indpendeent variable (weight).since the slope is 0.0976, so it is the value by which height would be change/increase when one unit of weight will change/ increase.

(c) for weight=150, height=52.92+0.0976*150=67.56

for weight=175, height=52.92+0.0976*175=70

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.