3) Assume that men\'s weights are normally distributed with a mean of 187.7 poun

Question

3) Assume that men's weights are normally distributed with a mean of 187.7 pounds and a standard deviation of 38.6 pounds. a) lf 1 man is randomly selected, what is the probability that his weight is between 175 and 190 pounds. b) If 10 men are randomly selected, what is the probability that their average weight is between 175 and 190 pounds? ) If a man's doctor tells him he is at the 45th percentile for weight, how much does he of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. 9 10) To find the standard deviation of the diameter of wooden dowels, the manufacturer measures 17 randomly selected dowels and finds the standard deviation of the sample to be 0.012 inches. He knows from long experience that the dowel's diameters are approximately normally distributed. Find the 99% confidence interval for the population standard deviation, , of all EXTRA CREDIT Two different tests are designed to measure employee productivity and dexterity. Several employees are randomly selected and tested with these results. Productivity 23 2528 21 21 25 26 30 34 36 Dexterity 49 53 59 42 47 53 55 63 67 75 a) Create a scatterplot of the data. From the scatterplot, describe the likely relationship between the two variables that were measured on these employees. b) Calculate r, the coefficient of linear correlation. What does the value of r tell you about the relationship between the two variables? Test the claim that these two variables are linearly correlated, to a 0.05 level of significance. c) d) If it makes sense to do so, calculate the equation of the line of linear regression. What is the slope of the line? How does it relate to the relationship between the tw variables? A new employee is hired and tests at a dexterity score of 64. What productivity score can we expect from this employee, according to the model we have created? e)

Explanation / Answer

3)

a)mean = mu = 187.7

sd =38.6

P(X'>175 to X'<190)

as we know z= (X'- mu)/(sd/sqrt(n))

where 1 man is randomly selected so n=1

z = (175-187.7)/(38.6) = -0.3290155 and so looking at the z table for -0.33 we get

Prob as = 0.3707

now for X'=190

z= (190-187.7)/36.6 =0.05958549 and so looking at the z table for 0.06 we get  0.5239222

so the prob = 0.5239222-0.3707 = 0.1532

b) now doing the same process for n=10

z = (175-187.7)/((38.6)/sqrt(10)) = -1.040439 and so looking at the z table for -1.04 we get

Prob as = 0.149068

now for X'=190

z= (190-187.7)/(36.6/sqrt(10)) =0.1987224and so looking at the z table for 0.1987224 we get  0.57876

so the prob = 0.57876-0.149068 = 0.429692

c) so to find the wt at 45th %tile we need z score at 0.45 value from the table and look for 0.45 from the area and find respective z value we get

z=-0.1256613

solving

z=(X'-mu)/(sd) (since we want a man's prob n=1 here)

-0.1256613 = (X'-187.5)/38.6

so., X'=182.6495 pounds

Ans = 182.6495 pounds at 45th percentile

As per the chegg policy we are allowed to answer 1 question at a time and so I have answered all the subparts of 1st question. If you want the answer for the subsequent questions pls post them seperately. Hope you understand.

Hope the above ans has helped you in understanding the problem pls upvote if it has really helped you. Good Luck!!

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