1. s pts) Do this one by suppose we measured the were normally men and found tha

Question

1. s pts) Do this one by suppose we measured the were normally men and found that the were distri height of 10,000 distri data with a mean inches and deviation of 4.0 inches. Answer the following questions and show your work: A. proportion of men a standard inches? What can be expected to have heights less than: 66, 70, 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 10,000) can be expected to have heights between: 65 and 71 inches, 72 and 75 inches? E. What height corresponds to the 4 height to the 55 height corresponds the 99th percentile? (If necessary, round the numbers, but keep at least two decimal places.) 2. (10 pts) Suppose for the same 10,000 men, we also measured their weight. Suppose the data were again normally height distributed. The average weight is 175 lbs., and the standard deviation is 25.0 lbs. Suppose the correlation between and weight is 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 B. variable and weight the X variable) variables: that is Write sentence or two that says what the equation of the line tells you about the relation between the when weight increases by one pound, how much does the predicted value of 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

1. n = sample size = 10000

Mean = 70

standard deviation = 4

X:Height of men

By using excel we can solve this question.

A. We have to find proportion of men expected to have heights less than 66,70,72,75 inches.

That is we have to find

P(X < 66) , P(X < 70), P(X < 72), P(X < 75)

Use command in excel = NORMDIST(x,mean,standard deviation,1)

We get

P(X < 66) = 0.1587

P(X < 70)= 0.5

P(X < 72)=0.6915

P(X < 75)=0.8944

B) Now we have to find P(X>64) , P(X>66), P(X>73),P(X>78)

Use command in excel = 1 - NORMDIST(x,mean,standard deviation,1)

We get

P(X>64)=0.9332

P(X>66)=0.8413

P(X>73)=0.2266

P(X>78)=0.0228

C) We have to find P(65 < X < 75) and P(71 < X < 72)

P(65 < X < 75) = P(X < 75) - P(X < 65 ) = 0.7887

P(71 < X < 72) = P(X < 72) - P(X < 71) = 0.0928

D) For finding number of men first we have to find P(65 < X < 71) and P(72 < X < 75)

P(65 < X < 71) = P(X < 71) - P(X < 65) = 0.4931

P(72 < X < 75) = P(X < 75) - P(X < 72)=0.2029

So number of men between 65 and 71 inches are 0.4931*10000 = 4931

number of men between 72 and 75 inches are 0.2029*10000 = 2029

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