In the library on a university campus, there is a sign in the elevator that indi

Question

In the library on a university campus, there is a sign in the elevator that indicates a limit of 18 persons. In addition, there is a weight limit of 2500 pounds. Assume that the average weight of students facility, and staff on campus is 158 pounds, that the standard deviation is 25 pounds and that the distribution of weights of individuals an campus is approximately normal. Suppose a random sample of 18 persons from the campus will be selected. (a) What is the mean of the x sampling distribution? (b) What is the standard deviation of the x sampling distribution? (Enter your answer to two decimal places.) (c) What average weights for a sample or 16 people will result in the total weight exceeding the weight limit or 2500 pounds? (Enter your answer to two decimal places.) (d) What is the chance that a random sample of 16 people will exceed the weight limit? (Round your answer to four decimal places.) You may need to use the appropriate table In Appendix A to answer this question. Suppose that the mean value of interpapillary distance (the distance between the pupils or the left and right eves) for adult males is 65 mm and that the population standard deviation is 5 mm. (a) If the distribution of interpapillary distance is normal and a random sample of n - 25 adult males is to hr selected. what is the probability that the sample mean distance x for theses 25 will be between 64 and 66 mm? (Round all your intermediate calculations to four decimal places. Round the answers to four decimal places.) What is the probability that the sample mean distance x for these 25 will he at least 87 mm? (b) Suppose that a sample of 100 adult males is to be obtained, without assuming that interpapillary distance is normally distributed. what is the approximate probability that the sample mean distance will be between 64 and 66 mm? (Round all your intermediate calculations to four decimal places. Round the answers to four decimal places) without assuming that interplay distance is normally distributed, what is the approximate probability that the sample mean distance will be at least 67 mm?

Explanation / Answer

Solution:-

7)
a. E(Xbar) = 156lbs
b. SD(Xbar) = 25/Sqrt16 = 6.25lbs
c. x-bar> 156.25
d. P(x-bar> 156.25) = P(Z > (156.25-156)/6.25) =P(Z > 0.04) = 0.4840

8)
a) P(63 < x-bar < 67) =?

z = (64 - 65)/(5/sqrt(25))= -1

z = (66 - 65)/(5/sqrt(25))= 1

P(64 < x-bar < 66) = P(-1 < z < 1)

= 0.6826

b) P(64 < x-bar < 66) =?

z = (64 - 65)/(5/sqrt(100))= -2

z = (66 - 65)/(5/sqrt(100))= 2

=> P(64 < x-bar < 66) = P(-2 < z < 2)

= 0.9544

=> P (67 < x) = P (4 < z) = 0

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