Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores calculated.
The average commute time via train from the Chicago O'Hare Airport to downtown is 60 minutes with a standard deviation of 15 minutes. Assume that the commute times are normally distributed. What proportion of commutes would be:
longer than 80 minutes?
less than 50 minutes?
between 45 and 75 minutes?
2 answers
a. Longer than 80 minutes?
z(80) = (80-60)/15 = 4/3
P(x> 80) = P(z> 4/3) = normal cdf(4/3,100) = 0.0912
---------------------------------------------------------
b. Less than 50 minutes?
z(50) = (50-60)/15 = -2/3
P(x< 50) = P(z< -2/3) = normal cdf(-100,-2/3) = 0.2525
----------------------------------------
c. Between 45 and 75 minutes?
z(45) = (45-60)/15 =
z(75) = (75-60)/15 =
P (x=60) = P(z= ) = normal cdf (-100, ) =
----------------------
?
z(80) = (80-60)/15 = 4/3
P(x> 80) = P(z> 4/3) = normal cdf(4/3,100) = 0.0912
---------------------------------------------------------
b. Less than 50 minutes?
z(50) = (50-60)/15 = -2/3
P(x< 50) = P(z< -2/3) = normal cdf(-100,-2/3) = 0.2525
----------------------------------------
c. Between 45 and 75 minutes?
z(45) = (45-60)/15 =
z(75) = (75-60)/15 =
P (x=60) = P(z= ) = normal cdf (-100, ) =
----------------------
?
0 answers