Among U.S. cities with a population of more than 250,000 the mean one-way commut

Question

Among U.S. cities with a population of more than 250,000 the mean one-way commute time to work is 24.3 minutes. The longest one-way travel time is in New York City, where the mean time is 38.2 minutes. Assume the distribution of travel times in New York City follows the normal probability distribution and the standard deviation is 7.0 minutes.

a.What percent of the New York City commutes are for less than 28 minutes? (Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places.)

  Percent %

b.What percent are between 28 and 33 minutes? (Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places.)

  Percent %

c.What percent are between 28 and 46 minutes? (Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places.)

  Percent %

Explanation / Answer

Mean ( u ) =38.2
Standard Deviation ( sd )=7
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P(X < 28) = (28-38.2)/7
= -10.2/7= -1.4571
= P ( Z <-1.4571) From Standard Normal Table
= 0.0725                  
7.25% are less than 28
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 28) = (28-38.2)/7
= -10.2/7 = -1.4571
= P ( Z <-1.4571) From Standard Normal Table
= 0.07254
P(X < 33) = (33-38.2)/7
= -5.2/7 = -0.7429
= P ( Z <-0.7429) From Standard Normal Table
= 0.22878
P(28 < X < 33) = 0.22878-0.07254 = 0.1562                  
15.62% are between 28 to 33
c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 28) = (28-38.2)/7
= -10.2/7 = -1.4571
= P ( Z <-1.4571) From Standard Normal Table
= 0.07254
P(X < 46) = (46-38.2)/7
= 7.8/7 = 1.1143
= P ( Z <1.1143) From Standard Normal Table
= 0.86742
P(28 < X < 46) = 0.86742-0.07254 = 0.7949                  
79.49% are between 28 to 46

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