Question

In response to the increasing weight of airline passengers, the Federal Aviation Administration told airlines to assume that passengers average 190 pounds in the summer, including clothing and carry-on baggage. But passengers vary: the FAA gave a mean but not a standard deviation. A reasonable standard deviation is 35 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commuter plane carries 25 passengers. What can we expect the sampling distribution for the mean weight of 25 passengers to be? Bimodal with μ = 190 and σ = 7. Normal with μ = 190 and σ = 7. Normal with μ = 190 and σ = 35. Bimodal with μ = 190 and σ = 35.

a.What is the probability, to four decimal places, that the average weight of 25 passengers will be less than 175 pounds?

b.What is the weight, to the nearest pound, below which only 10% of all 25 passenger samples have as their average weight? pounds

c.What is the weight, to the nearest pound, above which only 20% of all 25-passenger-samples have as their average weight? pounds

d.If the total weight of the 25 passengers is 5200 pounds, what is the average weight? pounds

e.What is the probability, to four decimal places, that the total weight of 25 passengers will exceed 5200 pounds?

Answer 1

std error=σx̅=σ/√n=35/√25 =

7.0000

Normal with μ = 190 and σ = 35

a_)

probability =P(X<175)=(Z<(175-190)/7)=P(Z<(-2.1429)=0.0161

b)

for 10th percentile critical value of z=-1.28

therefore corresponding value=mean+z*std deviation=181 pounds

c)

for 80th percentile critical value of z=0.84

therefore corresponding value=mean+z*std deviation=196 pounds

e_)

P( total weight of 25 passengers will exceed 5200 pounds ) =P(average weight exceeds 5200/25

=208):

probability =P(X>208)=P(Z>(208-190)/7)=P(Z>2.57)=1-P(Z<2.57)=1-0.9949=0.0051