Answer:
a. 0.0018 = 0.18% probability that your car has a worse highway gas mileage than the Civic
b. 89.06% of cars are likely to have a better city gas mileage than the Continental
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
a. The Honda Civic had a rating of 45 highway MPG. What is the probability that your car has a worse highway gas mileage than the Civic?
For Highway, we have that [tex]\mu = 29.09, \sigma = 5.46[/tex]
This probability is the pvalue of Z when X = 45. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{45 - 29.09}{5.46}[/tex]
[tex]Z = 2.91[/tex]
[tex]Z = 2.91[/tex] has a pvalue of 0.9982
1 - 0.9982 = 0.0018
0.0018 = 0.18% probability that your car has a worse highway gas mileage than the Civic.
b. The Lincoln Continental had a rating of 17 city MPG. What percent of cars are likely to have a better city gas mileage than the Continental?
City means that [tex]\mu = 22.37, \sigma = 4.37[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 17. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{17 - 22.37}{4.37}[/tex]
[tex]Z = -1.23[/tex]
[tex]Z = -1.23[/tex] has a pvalue of 0.1094
1 - 0.1094 = 0.8906
0.8906 = 89.06% of cars are likely to have a better city gas mileage than the Continental
