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epp P/x-75 10 P(Z = -1/15) - P(Z = -1.5) P(X²60) = 0·0013 (x²60) = 0.0668 L 60-75) Suppose that the wrapper of a certain candy bar lists its weight as 2.13 ounces. Naturally, the weights of individual bars vary somewhat. Suppose that the weights of these candy bars vary according to a normal distribution with mean u = 2.2 ounces and standard deviation o= 0.04 ounces. a. What proportion of candy bars weigh less than the advertised weight? b. What proportion of candy bars weigh more than 2.25 ounces? c. What proportion of candy bars weigh between 2.2 and 2.3 ounces? d. If the manufacturer wants to adjust the production process so that only 1 candy bar in 1000 weighs less than the advertised weight, what should the mean of the actual weights be (assuming that the standard deviation of the weights remains 0.04 ounces)? that weights of

epp P/x-75 10 P(Z = -1/15) - P(Z = -1.5) P(X²60) = 0·0013 (x²60) = 0.0668 L 60-75) Suppose that the wrapper of a certain candy bar lists its weight as 2.13 ounces. Naturally, the weights of individual bars vary somewhat. Suppose that the weights of these candy bars vary according to a normal distribution with mean u = 2.2 ounces and standard deviation o= 0.04 ounces. a. What proportion of candy bars weigh less than the advertised weight? b. What proportion of candy bars weigh more than 2.25 ounces? c. What proportion of candy bars weigh between 2.2 and 2.3 ounces? d. If the manufacturer wants to adjust the production process so that only 1 candy bar in 1000 weighs less than the advertised weight, what should the mean of the actual weights be (assuming that the standard deviation of the weights remains 0.04 ounces)? that weights of

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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can someone help with just number 8 please? Thank you very much.
of Zeddes' scores is 5. a. With which professor is a score of 90 more impressive? Support your answer with appropriate probability calculations and with a sketch. zoddes b. With which professor is a score of 60 more discouraging? Again support your answer B(x460) = P(X²-²7²260- P(X² 600)=0.00135 - with appropriate probability calculations and with a sketch. P (x-75 60-75 P(Z = -16) = P(ZZ-1.5) 17/(x²60) = 0.0668 8. Suppose that the wrapper of a certain candy bar lists its weight as 2.13 ounces. Naturally, the weights of individual bars vary somewhat. Suppose that the weights of these candy 10 10 bars vary according to a normal distribution with mean u = 2.2 ounces and standard deviation o= 0.04 ounces. a. What proportion of candy bars weigh less than the advertised weight? b. What proportion of candy bars weigh more than 2.25 ounces? c. What proportion of candy bars weigh between 2.2 and 2.3 ounces? d. If the manufacturer wants to adjust the production process so that only 1 candy bar in 1000 weighs less than the advertised weight, what should the mean of the actual weights be (assuming that the standard deviation of the weights remains 0.04 ounces)? Sample data from the National Center for Health Statistics reveal that weights of American men aged 20-29 have a mean of about 175 pounds and a standard deviation of about 35 pounds. For women the mean is about 140 pounds and the standard deviation is about 30 pounds. a. If these distributions are roughly normal, what percentage of men would you expect to weigh less than 150 pounds? Less than 200 pounds? Less than 250 pounds? Answer a. for women. 23.75% b. 25% c. Sample data from the l 97.72% 99.9740 -98,39% ne National Center for Health Statistics reveal that the anges are 29,0%, 82.1%, and 96.2% for men, 136-140 30 =-- -156-125) 35 0.2375 4200-175 = 0.7625 = 76.25% observed percentages in these
Transcribed Image Text:of Zeddes' scores is 5. a. With which professor is a score of 90 more impressive? Support your answer with appropriate probability calculations and with a sketch. zoddes b. With which professor is a score of 60 more discouraging? Again support your answer B(x460) = P(X²-²7²260- P(X² 600)=0.00135 - with appropriate probability calculations and with a sketch. P (x-75 60-75 P(Z = -16) = P(ZZ-1.5) 17/(x²60) = 0.0668 8. Suppose that the wrapper of a certain candy bar lists its weight as 2.13 ounces. Naturally, the weights of individual bars vary somewhat. Suppose that the weights of these candy 10 10 bars vary according to a normal distribution with mean u = 2.2 ounces and standard deviation o= 0.04 ounces. a. What proportion of candy bars weigh less than the advertised weight? b. What proportion of candy bars weigh more than 2.25 ounces? c. What proportion of candy bars weigh between 2.2 and 2.3 ounces? d. If the manufacturer wants to adjust the production process so that only 1 candy bar in 1000 weighs less than the advertised weight, what should the mean of the actual weights be (assuming that the standard deviation of the weights remains 0.04 ounces)? Sample data from the National Center for Health Statistics reveal that weights of American men aged 20-29 have a mean of about 175 pounds and a standard deviation of about 35 pounds. For women the mean is about 140 pounds and the standard deviation is about 30 pounds. a. If these distributions are roughly normal, what percentage of men would you expect to weigh less than 150 pounds? Less than 200 pounds? Less than 250 pounds? Answer a. for women. 23.75% b. 25% c. Sample data from the l 97.72% 99.9740 -98,39% ne National Center for Health Statistics reveal that the anges are 29,0%, 82.1%, and 96.2% for men, 136-140 30 =-- -156-125) 35 0.2375 4200-175 = 0.7625 = 76.25% observed percentages in these
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