Answer:
a) P(X < 30) = 0.0392.
b) P(28 < X < 32) = 0.2760
c) P(X > 35) = 0.1190
d) P(X > 31) = 0.8810
e) At least 35.7965 mpg
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
![\mu = 33, \sigma = 1.7](https://tex.z-dn.net/?f=%5Cmu%20%3D%2033%2C%20%5Csigma%20%3D%201.7)
a. P(X<30)
This is the pvalue of Z when X = 30. So
![Z = -1.76](https://tex.z-dn.net/?f=Z%20%3D%20-1.76)
has a pvalue of 0.0392.
Then
P(X < 30) = 0.0392.
b) P(28 < X < 32)
This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 28. So
X = 32
![Z = -0.59](https://tex.z-dn.net/?f=Z%20%3D%20-0.59)
has a pvalue of 0.2776.
X = 28
![Z = -2.94](https://tex.z-dn.net/?f=Z%20%3D%20-2.94)
has a pvalue of 0.0016.
0.2776 - 0.0016 = 0.2760.
So
P(28 < X < 32) = 0.2760
c) P(X>35)
This is 1 subtracted by the pvalue of Z when X = 35. So
![Z = 1.18](https://tex.z-dn.net/?f=Z%20%3D%201.18)
has a pvalue of 0.8810.
1 - 0.8810 = 0.1190
So
P(X > 35) = 0.1190
d. P(X>31)
This is 1 subtracted by the pvalue of Z when X = 31. So
![Z = -1.18](https://tex.z-dn.net/?f=Z%20%3D%20-1.18)
has a pvalue of 0.1190.
1 - 0.1190 = 0.8810
So
P(X > 31) = 0.8810
e. the mileage rating that the upper 5% of cars achieve.
At least the 95th percentile.
The 95th percentile is X when Z has a pvalue of 0.95. So it is X when Z = 1.645. Then
![X - 33 = 1.645*1.7](https://tex.z-dn.net/?f=X%20-%2033%20%3D%201.645%2A1.7)
![X = 35.7965](https://tex.z-dn.net/?f=X%20%3D%2035.7965)
At least 35.7965 mpg