Question

According to the Environmental Protection Agency (EPA), the 2018 Toyota Camry L drives an average of 420.5 miles on a full tank of gas. Assume the mileage follows a normal distribution with a standard deviation of 50 miles. Answer the following questions:

17.) Determine the number of miles that the car will travel with 90%, 30%, and 50% probability on a full tank of gas.

18.) Determine the UPPER AND LOWER value(s) for the interval of miles traveled on a full tank of gas around the mean that includes approximately 68% of miles for this car.

19.) Determine the UPPER AND LOWER value(s) for the interval of miles traveled on a full tank of gas around the mean that includes approximately 95% of miles for this car.

20.) Determine the UPPER AND LOWER value(s) for the interval of miles traveled on a full tank of gas around the mean that includes approximately 99.7% of miles for this car.

Answer 1

for normal distribution z score =(X-μ)/σx
here mean=       μ=	420.5
std deviation   =σ=	50.000

17)

for top 90% values, critical z =-1.28

therefore corresponding miles=mean+z*std deviation=

356.50 miles

for top 30% values, critical z =0.52

corresponding miles=mean+z*std deviation=446.50

for top 50% values, critical z =0

corresponding miles=mean+z*std deviation=420.5

18)

for 68% middle values are 1 standard deviation from mean:

UPPER value =mean+1 standard deviation =420.5+50 =470.5

LOWER value =mean-1 standard deviation =420.5-50 =370.5

19)

for 95% middle values are 2 standard deviation from mean:

UPPER value =mean+2* standard deviation =420.5+2*50 =520.5

LOWER value =mean-2* standard deviation =420.5-2*50 =320.5

20)

for 99.7% middle values are 3 standard deviation from mean:

UPPER value =mean+3* standard deviation =420.5+3*50 =270.5

LOWER value =mean-3* standard deviation =420.5-3*50 =570.5