Question

1. (20 pts) Do this one by hand. Suppose we measured the height of 5,000 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 questions using Table A and show your work:

1. What proportion of men can be expected to have heights less than 66 inches? Less than 75 inches? .1587 and .8944
2. What proportion of men can be expected to have heights greater than 64 inches? Greater than 73 inches? .9332 and .2266
3. What proportion of men can be expected to have heights between 65 and 75 inches? Between 71 and 72 inches? .7888 and .09982
4. What height corresponds to the 43^rd percentile? 69.296 What height corresponds to the 55^th percentile? 70.504 What height corresponds to the 99^th percentile? 79.304 (If necessary, round the numbers, but keep at least two decimal places.)

2. (15 pts) Suppose for the same 5,000 men, we also measured their weight. Suppose the data were again normally distributed. The average weight is 160.0 lbs., and the standard deviation is 20.0 lbs. Suppose the correlation between height and weight is r = +.61. Use the ‘shortcut’ formulas to answer the questions and show your work:

1. What is the slope of the best-fit line to predict height from weight? 0.122 What is the intercept of the line? 50.48 (Make height the Y variable and weight the X variable).
2. Write the full equation of the line to predict height from weight. Use variable names and numbers for the coefficients to make the equation as complete and precise as possible. Height=0.122 x Weight + 50.48
3. 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?
4. What is your best guess for the height of a man who weighs 150 lbs.? 68.78 inches What is your best guess for the height of a man who weighs 160 lbs.? 70 inches

3. (20 pts) For the same data as in Questions 1-2, make a detailed drawing by hand of what the scatterplot would look like. (You don’t have the original data, but you can actually provide quite a bit of information about the look of the scatterplot!) Be sure to clearly indicate each of the following: which variable is X or Y, the range on the X and Y axis of each of the variables (you can figure out the approximate range that will include most of the scores of height and weight by knowing the means and standard deviations, taking into account the sample size, and by using Table A in Appendix D), the equation of the best-fit line for predicting Y from X, and a sense of the dispersion of points that is close to a correlation of +.61 (see examples in chapter 6). Be sure to accurately draw the regression line on the scatterplot, label it with its equation (taken from Question 2), and also to include a few sample deviates for the best-fit line (the deviates will be vertical – again, see examples in chapter 6). (Note: You don’t need to draw all 5,000 data points – just include enough to give a sense of the spread of the data for the given value of the Pearson r.)

Question # 3 please

Answer 1

we found that the data of 5000 men are normally distributed with mean 70.0 inches and standard deviation 4.0 inches

let X be the random variable and x follows(70,4²)

let z =(x-70)/4 follows normal (0,1)

(A) We are here to find the proportion of men can be expected to have heights less than 66 inches

i,e, P(x<65)=P(((x-70)/4)<((65-70)/4)

                  =P(z<-1)=1-cap phi(1)=1-0.8413447=0.1586 (by normal table)

to find the proportion of men can be expected to have heights less than 75 inches

i.e,

P(x<75)=P(((x-70)/4)<((75-70)/4)

                  =P(z<1.25)=cap phi(1.25)=0.8943502 (by normal table)

(B) to find proportion of men can be expected to have height more than 64 inches

i.e,

P(x>64)=P(((x-70)/4)>((64-70)/4)=1-P(((x-70)/4)<((64-70)/4)

                  =1-P(z<-1.5)=cap phi(1.5)=0.9331928 (by normal table)

to find proportion of men can be expected to have height more than 73 inches

P(x>73)=P(((x-70)/4)>((73-70)/4)=1-P(((x-70)/4)<((73-70)/4)

                  =1-P(z<0.75)=1-cap phi(0.75)=0.2266274 (by normal table)

(C) to find the proportion of men can be expected to have heights between 65 and 75 inches

i.e,

P(65<x<75)=P(((65-70)/4)<((x-70)/4)<((75-70)/4))

                    =P(-1.25<z<1.25)=cap phi(1.25)-cap phi (-1.25)

                                                =(2*0.8943502)-1=0.7887004

for height between 71 and 72 inches

i.e,

P(71<x<72)=P(((71-70)/4)<((x-70)/4)<((72-70)/4))

                    =P(0.25<z<0.5)=cap phi(0.5)-cap phi (0.25)

                                                =0.6914625-0.5987063=0.0927562