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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