In determining automobile-mileage ratings, it was found that the mpg (X) for a certain model is normally distributed, with a mean of 33 mpg and a standard deviation of 1.7 mpg. Find the following:. a. P(X<30)
b. P(28 c. P(X>35)
d. P(X>31)
e. the mileage rating that the upper 5% of cars achieve.

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:

a. P(X<30)

This is the pvalue of Z when X = 30. So

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

has a pvalue of 0.2776.

X = 28

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

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

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

At least 35.7965 mpg

for b its (t) and c its (d)

step-by-step explanation:

depends on how many days you are talking about

step-by-step explanation: