Question

question 20 (bonus) (1 point)
an automobile manufacturer introduces a new model that averages 22.0 miles per gallon in the city. a person who plans to purchase one of these new cars wrote the manufacturer for the details of the tests, and found out that the standard deviation is 3.8 miles per gallon. assume that in - city mileage is approximately normally distributed.
____% of the automobiles have between 18 and 21 miles per gallon.
round the z - score to two decimal places if needed. round your answer to two decimal places. do not include the % sign in your answer.
your answer:
answer
question 21 (bonus) (1 point)
the exam scores for a large class is approximated by a normal distribution with a mean of 78 and standard deviation of 13.2.
____% of the exam scores are between 80 and 90.
round the z - score to two decimal places if needed. round you answer to two decimal places. do not include the % sign in your answer.
your answer:
answer
question 22 (1 point)

Explanation:

Step1: Calculate z - scores for Question 20

For \(x = 18\), \(z_1=\frac{18 - 22}{3.8}=\frac{-4}{3.8}\approx - 1.05\)
For \(x = 21\), \(z_2=\frac{21 - 22}{3.8}=\frac{-1}{3.8}\approx - 0.26\)

Step2: Find probabilities from z - table

\(P(Z\lt - 1.05)=0.1469\), \(P(Z\lt - 0.26)=0.3974\)
\(P(-1.05\lt Z\lt - 0.26)=P(Z\lt - 0.26)-P(Z\lt - 1.05)\)
\(P(-1.05\lt Z\lt - 0.26)=0.3974 - 0.1469 = 0.2505\)

Step3: Calculate z - scores for Question 21

For \(x = 80\), \(z_1=\frac{80 - 78}{13.2}=\frac{2}{13.2}\approx0.15\)
For \(x = 90\), \(z_2=\frac{90 - 78}{13.2}=\frac{12}{13.2}\approx0.91\)

Step4: Find probabilities from z - table

\(P(Z\lt0.15)=0.5596\), \(P(Z\lt0.91)=0.8186\)
\(P(0.15\lt Z\lt0.91)=P(Z\lt0.91)-P(Z\lt0.15)\)
\(P(0.15\lt Z\lt0.91)=0.8186 - 0.5596=0.259\)

Answer:

Question 20: \(25.05\)
Question 21: \(25.90\)