2.
Assume that the readings on the thermometers are normally distributed with a mean of 0 and standard deviation of 1.00C. A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to P91?, the 91st percentile. This is the temperature reading separating the bottom 91% from the top
A survey found that women's heights are normally distributed with mean 63.6 in and standard deviation 2.2 in. A branch of the military requires wo men's heights to be between 58 in and 30 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. if this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements? Click to View page 1 of the table. Click to View page 2 of the table. a. The percentage of women who meet the height requirement is |:|%. (Round to two decimal places as needed.) Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? 0 A. Yes, because the percentage of women who meet the height requirement is fairly large. 0 B. No, because the percentage ofwomen who meet the height requirement is faiy small. 0 C. Yes, because a large percentage of women are not allowed to join this branch of the military because of their height. 0 D. No, because only a small percentage of women are not allowed to join this branch of the military because of their height. b. For the new height requirements, this branch of the military requires women's heights to be at least in and at most in. (Round to one decimal place as needed.) The weights of a certain brand of candies are non'nally distn'buted with a mean weight ofIJBSQB g and a standard deviation otl].[152 g. Asample of these candies came from a package containing 466 candies, and the package 39 label stated that the net weight is 39? 9 g. (If every package has 4-66 candies. the mean weight ofthe candies must exceed 466 = I] 8538 9 For the net contents to weigh at least 397 E g } a. it 1 candy is randomly selected, nd the probability that it weighs more than 0.8533 g The probability is . (Round to four decimal places as needed.) b. It 456 candies are randomly selected. nd the probability that their mean weight is at least 0.8538 g. The probability that a sample of 466 candies will have a mean of 0.3533 g or greater is (Round to four decimal places as needed.) c. Given these results. does it seem that the candy company is providing consumers with the amount claimed on the label? " because the probability of getting a sample mean of 0.8538 g or greater when 456 candies are selected Y exceptionally small. Assume that the readings on the thermometers are normally dialn'buled with a mean ol0 and standard deviation of1.00C.A lhennomeler is randomly selected and tested. Draw a sketch and nd the temperature reading corresponding to P91, the B15t percentile, Th'E is the temperature reading separating the bottom 91% from the top 9% Click to View page I of the table Clid
