Question

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.2 minutes and a standard deviation of 3.5 minutes. Assume that the distribution of taxi and takeoff times is approximately normal. You may assume that the jets are lined up on a runway so that one taxies and takes off immediately after the other, and that they take off one at a time on a given runway.

 (a) What is the probability that for 39 jets on a given runway, total taxi and takeoff time will be less than 320 minutes? (Round your answer to four decimal places.) 

 (b) What is the probability that for 39 jets on a given runway, total taxi and takeoff time will be more than 275 minutes? (Round your answer to four decimal places.)

 (c) What is the probability that for 39 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes? (Round your answer to four decimal places.)

A European growth mutual fund specializes in stocks from the British Isles, continental Europe, and Scandinavia. The fund has over 500 stocks. Let x be a random variable that represents the monthly percentage return for this fund. Suppose x has mean μ = 1.3% and standard deviation σ = 1.5%.

Answer 1

1)

for normal distribution z score =(X-μ)/σx
mean μ=	8.2
standard deviation σ=	3.5

std error=σx̅=σ/√n=

0.5604

a)

probability =P(X<320/39)=(Z<(8.21-8.2)/0.56)=P(Z<(0.0092)=0.5037

b)

probability =P(X>275/39)=P(Z>(7.05-8.2)/0.56)=P(Z>-2.05)=1-P(Z<-2.05)=1-0.0202=0.9798

c)

probability =P(275/390.4838

2)

b)

std error=σx̅=σ/√n=

0.5000

probability =P(10.6449

c)

std error=σx̅=σ/√n=

0.3536

probability =P(10.7784

e)

probability =P(X>2)=P(Z>(2-1.3)/0.354)=P(Z>1.98)=1-P(Z<1.98)=1-0.97614=0.0239