JustAnswer
#1 Woman’s heights are normally distributed with mean 63.6 in. and standard deviation of 2.5 in. A social organization for tall people has a requirement that women must be at least 69 in. tall. What percentage of women meets that requirement?
|z = (x - μ)/σ = (69 - 63.6)/2.5 = 2.16 |
|P(x > 69 inches) = P(z > 2.16) = 0.0154 |
|1.54% of the women meet the requirement |
#4 Assume the human body temperatures are normally distributed with a mean of 98.21 F and a standard deviation of 0.63 F.
a. A hospital uses 100.6º F as the lowest temperature considered to be a fever. What percentage of normal and healthy persons would be considered to have a fever? Does this percentage suggest that a cutoff of 100.6º F is appropriate?
b. Physicians want to select a minimum temperature for requiring further medical tests. What should that temperature be, if we want only 5.0% of healthy people to exceed it? (such a result is a false positive, meaning that the test result is positive, but the subject is not really sick)
|(a) z = (x - μ)/σ = (100.6 - 98.21)/0.63 = 3.7937 |
|P(x > 100.6 F) = P(z > 3.7937) = 0.0001 |
|About 0.01% of normal and healthy persons would be considered to have a fever. Since this value is very small, the cut-off is appropriate. |
|(b) Right-tail z- score for 5% is 1.6449 |
|x = μ + z * σ = 98.21 + 1.6449 * 0.63 = 99.25 |
|The temperature should be 99.25 F |
#6 An airliner carries 50 passengers and have doors with a height of 76 in. Heights of men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches.
a. If a male passenger is randomly selected …find the probability that he can fit through the doorway without bending.
b. If half of the passengers are men, find the probability that the mean height of the 25 men is less than 76 inches.
c. When considering the comfort and safety of passengers, which result is more relevant the probability from part (a) or the probability from part (b)? Why?
d. When considering the comfort and safety of passengers, why are women ignored in this case?
|(a) z = (x - μ)/σ = (76 - 69)/2.8 = 2.5 |
|P(x < 76 inches) = P(z < 2.5) = 0.9938 |
|The probability is 0.9938 |
|(b) z = (x-bar - μ)/(σ/√n) = (76 - 69)/(2.8/√25) = 12.5 |
|P(x-bar < 76 inches) = P(z < 12.5) = 1.0 |
|The probability is 1.0 |
|(c) The result from (a) is more relevant since it shows the probability that a random male passenger will comfortably pass through the door. |
|(d) Generally women’s heights are less than men’s heights. If doors are designed for men’s heights, they will meet the requirements of women also. That’s why |
|women are ignored. |
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