Answer:
The probability that a car selected at random has
(a) less than 26 mpg is 0.0668
(b) greater than 34 mpg is 0.3085
(c) between 30 and 34 mpg is 0.383
Step-by-step explanation:
The question is missing, complete question is below:
A certain model of automobile has its gas mileage (in miles per gallon, or mpg) normally distributed, with a mean of 32 mpg and a standard deviation of 4 mpg. Find the probability that a car selected at random has the following gas mileages.
(a) less than 26 mpg (b) greater than 34 mpg (c) between 30 and 34 mpg
z-score of a <em>specific </em>value in the distribution of gas mileages can be found using:
z= where
- X is the specific value
- M is the mean gas mileage in mpg (32 mpg)
- s is the standard deviation of the distribution (4 mpg)
Then
a) P(X<26) = P(z<z*) where
z*= =-1.5
P(z<z*) = 0.0668
b) P(X>34)=1-P(x<34)=1-P(z<z*) where
z*= =0.5
Then 1-P(z<z*) = 1-0.6915 = 0.3085
c) P(30<X<34) =P(X<34)-P(X<30)
P(X<30)=P(z<z*) where
z*= =-0.5
Then P(z<z*)=0.3085 Since P(X<34)=0.6915 we have
P(30<X<34) =0.6915-0.3085 =0.383
