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## QNT 351 Week 4, Individual Assignment - MyStatsLab

by: Experthelper Notetaker

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# QNT 351 Week 4, Individual Assignment - MyStatsLab

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QNT 351 Week 4, Individual Assignment - MyStatsLab
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Date Created: 11/09/15
Week 4 Problems 1. Determine whether the random variable is discrete or continuous. a. the number of textbook authors now sitting at a computer b. the weight of a T-bone steak c. the number of fish caught during a fishing tournament d. the time it takes to fly from city A to city B e. the number of points scored during a basketball game a. Is the number of textbook authors now sitting at a computer? A. The random variable is discrete B. The random variable is continuous b. Is the weight of a T-bone steak discrete or continuous A. The random variable is continuous B. The random variable is discrete c. Is the number of fish caught during a fishing tournament discrete or continuous? A. The random variable is continuous B. The random variable is discrete d. Is the time it takes to fly from city A to city B discrete or continuous? A. The random variable is discrete B. The random variable is continuous e. Is the number of points scored during a basketball game discrete or continuous? A. The random variable is discrete B. The random variable is continuous 2. Determine whether the random variable is discrete or continuous. In each case, state the possible values of the random variable. (a) the number of light bulbs that burn out in the next week in room with 17 bulbs (b) the amount of rain in City B during April (a) Is the number of light bulbs that burn out in the next week in room with 17 bulbs discrete or continuous? A. The random variable is continuous. The possible values are 0≤x≤17 B. The random variable is discrete. The possible values are 0, 1, 2, …, 17 C. The random variable is continuous. The possible values are 0, 1, 2, …, 17 D. The random variable is discrete. The possible values are 0≤x≤17 (b) Is the amount of rain in City B during April discrete or continuous? A. The random variable is discrete. The possible values are r=1, 2, 3, … B. The random variable is discrete. The possible values are r≥0 C. The random variable is continuous. The possible values are r≥0 D. The random variable is continuous. The possible values are r=1, 2, 3, … 3. The random variable x has the following discrete probability distribution. Complete parts a trough f. a. List the values x may assume. 1, 4, 5, 8, 9 b. What value of x is most probable? 9 c. Display the probability distribution as a graph. Choose the correct graph bellow. A. B. C. D. Correct Answer is D d. Find P(x=8) P(x=8)=.1 e. Find P(x≥5) P(x≥5)=.7 f. Find P(x>2) P(x>2)=.9 4. A discrete random variable x can assume five possible values 2, 3, 5, 7, and 9. Its probability distribution is shown here. Complete parts a trough c. a. What is p(5)? p(5) = .26 b. What is the probability that x equals 2 or 9? p(x=2 or x=9) = .41 c. What is p (x≤7)? p (x≤7) = .74 5. Toss three fair coins and let x equal the number of tails observed. a. Identify the sample points associated with this experiment and assign a value of x to each sample point b. Calculate p(x) for the values x=1 and x=2 c. Construct a graph for p(x) d. What is P(x=2 or x=3)? a. Choose the correct answer bellow A. x=0, 1, 2, 3 B. x=1, 2, 3 C. x=0, 1, 2 D. x=1, 2 b. Calculate p(x) for x=1 p(1) = .375 Calculate p(x) for x=2 p(2) = .375 c. Choose the correct graph bellow A. B. C. D. d. P(x=2 or x=3) = .5 6. Find the following probabilities for the standard normal random variable z. a. P(z>1.79) b. P(z< -1.46) c. P(0.54 ≤ z ≤ 2.67) d. P(- 2.34 < z > 1.77) a. P(z>1.79) = .0367 b. P(z< -1.46) = .0721 c. P(0.54 ≤ z ≤ 2.67) = .2908 d. P(- 2.34 < z > 1.77) = .952 7. Find the area under the standard normal probability distribution between the following pairs of z – scores. a. z=0 and z=3.00 b. z=0 and z=1.00 c. z=0 and z=2.00 d. z=0 and z=0.62 a. the area under the standard normal probability distribution is .4987 b. the area under the standard normal probability distribution is .3413 c. the area under the standard normal probability distribution is .4772 d. the area under the standard normal probability distribution is .2324 8. Find the area under the standard normal probability distribution between the following pairs of z – scores. a. z=0 and z=1.00 b. z=0 and z=2.00 c. z=0 and z=3.00 d. z=0 and z=0.62 a. the area under the standard normal probability distribution is .3413 b. the area under the standard normal probability distribution is .4772 c. the area under the standard normal probability distribution is .4987 d. the area under the standard normal probability distribution is .2324 9. Find the following probability for the standard normal random variable z. P( - 1.5 ≤ z ≤ 1.5) P( - 1.5 ≤ z ≤ 1.5) = .8664 10. Suppose the random variable x is best described by a normal distribution with µ = 26 and  = 5. Find the z score that corresponds to each of the following x – values. a. x=17 b. x=26 c. x=11 d. x=13 e. x=20 f. x=32 a. The z score that corresponds to x = 17 is z = - 1.8 b. The z score that corresponds to x = 26 is z = 0 c. The z score that corresponds to x = 11 is z = - 3 d. The z score that corresponds to x = 13 is z = - 2.6 e. The z score that corresponds to x = 20 is z = - 1.2 f. The z score that corresponds to x = 32 is z = 1.2 11. Suppose x is a normal distributed random variable with µ = 11 and  = 2. Find each of the following probabilities. a. P(x≥14.5) b. P(x≤9) c. P(12.56≤x≤16.12) d. P(5.84≤x≤13.42) a. P(x≥14.5) = .0401 b. P(x≤9)= .1587 c. P(12.56≤x≤16.12) = .2125 d. d. P(5.84≤x≤13.42) = .882 12. The ages of a group of 50 women are approximately normally distributed with a mean of 48 years and a standard deviation of 5 years. One woman is randomly selected from the group, and her age is observed. a. Find the probability that her age will fall between 55 and 59 years. b. Find the probability that her age will fall between 47 and 53 years. c. Find the probability that her age will be less than 35 years. d. Find the probability that her age will exceed 41 years a. the probability that her age will fall between 55 and 59 years is .0669 b. the probability that her age will fall between 47 and 53 years is 0.4206 c. the probability that her age will be less than 35 years is .0047 d. the probability that her age will exceed 41 years is .9192 13. Financial analysts who make forecasts of stock prices are categorized as either “buy-side” analysts or “sell-side” analysts. The mean and standard deviation of the forecast errors for both types of analysts are shown is the table to the right. Assume that the distribution of forecast errors are approximately normally distributed. a. Find the probability that a buy-side analyst has a forecast error of + 2.01 or higher b. Find the probability that a sell-side analyst has a forecast error of + 2.01 or higher a. the probability that a buy-side analyst has a forecast error of + 2.01 or higher is .2776 b. the probability that a sell-side analyst has a forecast error of + 2.01 or higher is .0066 14. The mean gas mileage for a hybrid car is 57 miles per gallon. Suppose that the gasoline mileage is approximately normally distributed with a standard deviation of 3.5 miles per gallon. a. What is the probability that a randomly selected hybrid gets more than 62 miles per gallon? b. What is the probability that a randomly selected hybrid gets 50 miles per gallon or less? c. What is the probability that a randomly selected hybrid gets between 57 and 62 miles per gallon? d. What is the probability that a randomly selected hybrid gets less than 46 miles per gallon? a. the probability that a randomly selected hybrid gets more than 62 miles per gallon is .0764 b. the probability that a randomly selected hybrid gets 50 miles per gallon or less is .0228 c. the probability that a randomly selected hybrid gets between 57 and 62 miles per gallon .4236 d. the probability that a randomly selected hybrid gets less than 46 miles per gallon .0008 15. Resource Reservation Protocol (RSVP) was originally designed to establish signaling links for stationary networks. RSVP was applied to mobile wireless technology. A simulation study revealed that the transmission delay (measured is milliseconds) of an RSVP linked wireless device has an approximate normal distribution with mean µ=46.5 milliseconds and  = 8.5 milliseconds. Complete part a and b. a. What is the probability that the transmission delay is less than 57 milliseconds? P(x<57) = .8925 b. What is the probability that the transmission delay is between 40 and 60 milliseconds? P(40≤x≤60) = .7205

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