Among US cities with a population of more than 250000 the me

Among U.S. cities with a population of more than 250,000 the mean one-way commute to work is 24.3 minutes. The longest one-way travel time is New York City, where the mean time is 38.5 minutes. Assume the distribution of travel times in New York City follows the normal probability distribution and the standard deviation is 7.2 minutes.

What percent of the New York City commutes are for less than 26 minutes? (Round the intermediate values to 2 decimal places. Round your answers to 2 decimal places. Omit the \"%\" sign in your response.)

What percent are between 26 and 34 minutes? (Round the intermediate values to 2 decimal places. Round your answers to 2 decimal places. Omit the \"%\" sign in your response.)

What percent are between 26 and 45 minutes? (Round the intermediate values to 2 decimal places. Round your answers to 2 decimal places. Omit the \"%\" sign in your response.)

Solution

Mean ( u ) =38.5
Standard Deviation ( sd )=7.2
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 26) = (26-38.5)/7.2
= -12.5/7.2= -1.7361
= P ( Z <-1.7361) From Standard Normal Table
= 0.0413 ~ 4.13%                  

b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 26) = (26-38.5)/7.2
= -12.5/7.2 = -1.7361
= P ( Z <-1.7361) From Standard Normal Table
= 0.04127
P(X < 34) = (34-38.5)/7.2
= -4.5/7.2 = -0.625
= P ( Z <-0.625) From Standard Normal Table
= 0.26599 ~ 26.599%

c)
P(26 < X < 34) = 0.26599-0.04127 = 0.2247                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 26) = (26-38.5)/7.2
= -12.5/7.2 = -1.7361
= P ( Z <-1.7361) From Standard Normal Table
= 0.04127
P(X < 45) = (45-38.5)/7.2
= 6.5/7.2 = 0.9028
= P ( Z <0.9028) From Standard Normal Table
= 0.81668
P(26 < X < 45) = 0.81668-0.04127 = 0.7754 ~ 77.54%