15. Given that z is a standard normal random variable, find z for each situation

Question

15. Given that z is a standard normal random variable, find z for each situation. a. b. c. d. e. Given that z is a standard normal random variable, find z for each situation. a. The area to the right of z is.01. b. The area to the right of z is.025 c. The area to the right of z is .05. d. The area to the right of z is .10. The area to the left of z is .2119. The area between -z and z is .9030. The area between -z and z is .2052. The area to the left of z is .9948. The area to the right of z is.6915. 16. Applications The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation Statistics website, November 2, 2012). Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $110. a. What is the probability that a domestic airfare is $550 or more? b. What is the probability that a domestic airfare is $250 or less? 17.

Explanation / Answer

Ans:

1)

a)P(Z<=z)=0.2119

z=normsinv(0.2119)

z=-0.80

b)

Area outside -z and z=1-0.9030=0.097

one tailed area=0.097/2=0.0485

z=normsinv(0.0485) or normsinv(1-0.0485)

z=1.66

c)

Area outside -z and z=1-0.2052=0.7948

one tailed area=0.7948/2=0.3974

z=normsinv(0.3974) or normsinv(1-0.3974)

z=0.26

d)

P(Z<=z)=0.9948

z=normsinv(0.9948)=2.56

e)

P(Z>z)=0.6915

P(Z<=z)=1-0.6915=0.3085

z=normsinv(0.3085)

z=-0.50

16)

a)P(Z>z)=0.01

P(Z<=z)=1-0.01=0.99

z=2.33

b)P(Z>z)=0.025

P(Z<=z)=1-0.025=0.975

z=1.96

c)P(Z>z)=0.05

P(Z<=z)=1-0.05=0.95

z=normsinv(0.95)

z=1.64

d)P(Z>z)=0.1

P(Z<=z)=1-0.1=0.9

z=normsinv(0.9)

z=1.28

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