Homework Set - Module 10

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Gloria Hansen Homework Set – Module 9 4.85) Find the following probabilities for the standard normal random variable z: a) P (z > 1.46) 1 – 0.9279 = .0721 b) P (z < -1.56) = .0594 c) P (.67 <= z < 2.41) .9920 - .7486 = .2434 d) P (-1.96 <= z < -.33) .3707 - .025 = .3457 e) P (z >= 0) 1 - .5 = .5 f) P (-2.33 < z < 1.50) .9332 - .0099 = .9233 4.88) Find a value of the standard normal random variable z, call it z0, such that: a) P (z >= z0) = .05 1 - .05 = .95 z0 = 1.64 b) P (z >= z0) = .025 1 – 0.25 = .975 z0 = 1.96 c) P (z <= z0) = .025 = -1.96 d) P (z >= z0) = .10 1 - .10 = .90 z0 = 1.28 e) P (z > z0) = .10 z0 = 1.28 4.91) Suppose the random variable x is best described by a normal distribution with a mean = 30 and a standard deviation = 4. Find the z-score that corresponds to each of the following x values:

Gloria Hansen a) X = 20 20-30/4 = -2.5 b) X = 30 30-30/4 = 0 c) X = 2.75 2.75-30/4 = -6.8125 d) X = 15 15-30/4 = -3.75 e) X = 35 35-30/4 = 1.25 f) X = 25 25-30/4 = -1.25 4.93) Suppose x is a normally distributed random variable with a mean = 11 and standard deviation = 2. Find each of the following: a) P (10 <= x <= 12) P (-.5 <= z <= .5) = .3829 b) P (6 <= x <= 10) P (-2.5 <= z <= -.5) = .3023 c) P (13 <= x <= 16) P (1 <= z <= 2.5) = .3624 d) P (7.8 <= x <= 12.6) P (-1.6 <= z <= .8) = .7333 e) P (x >= 13.24) P (z >= 1.12) = .1314 f) P (x >= 7.62) P (z >= -1.68) = .9535 4.94) Suppose x is a normally distributed random variable with mean = 50 and standard deviation = 3. Find a value of the random variable, call it x0, such that:

Gloria Hansen a) P (x <= x0) = .8413 50 + (1*3) = 53 b) P (x > x0) = .025 50 + (1.96*3) = 55.88 c) P (x > x0) = .95 50 + (1.64*3) = 54.92 d) P (41 <= x < x0) = .8630 = 50 + (1.1*3) = 53.3 e) 10% of the values of x are less than x0 P (z <= z0) = .10 1 - .10 = .90 = 50 + (1.28*3) = 53.84 f) 1% of the values of x are greater than x0 P (z <= z0) = .01 1 - .01 = .99 = 50 + (2.32*3) = 56.96 4.97) With a variable life insurance policy, the rate of return on the investment (i.e., the death benefit) varies from year to year. A study of these variable return rates was published in International Journal of Statistical Distributions (Vol. 1, 2015). A transformed ratio of the return rates (x) for two consecutive years was shown to have a normal distribution, with mean = 1.5 and standard deviation = .2. Use the standard normal table or statistical software to find the following probabilities. a) P (1.3 < x < 1.6) .1915 + .3413 = .5328 b) P (x > 1.4) = .6915 c) P (x < 1.5) = .5 4.99) Recall that a major apparel retailer has two distribution centers (DCs) to fulfill all its online orders – one located in the eastern United States and one in the western United States. Based on the study, we can assume that the delivery time x (in business days) for online orders fulfilled by the eastern DC is normally distributed with mean = 5.22 and standard deviation = .77. Similarly, assume that the delivery time x for online orders fulfilled by the western DC is normally distributed with mean = 6.95 and standard deviation = .55. a) For online orders filled by the eastern DC, fins P ( x <= 5).

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Gloria Hansen 5 – 5.22/.77 = .3859 b) For online orders filled by the eastern DC, find P (4 <= x <= 7) 2 – 5.22/.77 = .9345 c) For online orders filled by the eastern DC, find the value a, so that P (x > a) = .30. 5.22 + (.52 * .77) = 5.62 d) Repeat parts a-c for online orders filled by the western DC. (a) = (b) = (c) = (d) = 4.100) Recall that the level of support for corporate sustainability (measured on a quantitative scale ranging from 0 to 160 points) was obtained for each of 992 senior managers at CPA firms. The accompanying StatCrunch printout gives the mean and standard deviation for the level of support variable. It can be shown that level of support is approximately normally distributed. a) Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is less than 40 points. P (z < 40 – 67.755/26.87) P ( z < -1.03289) = .15151 b) Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is between 40 and 120 points. P (40 – 67.755/26.871 <= z <= 120-67.755/26.871) P (-1.03289 <= z <= 1.94428) = .97391 - .15151 = .8223 c) Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is greater than 120 points. P (z < 120 – 67.755/26.87) P ( z < 1.94428) 1 – 0.9738 = .0262 d) One-fourth of the 992 senior managers indicated a level of support for corporate sustainability below what value? .25 = x-67.755/26.871 X = 6.71775 + 67.755 X = 74.47275

Gloria Hansen

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