Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 33, \sigma = 1.7[/tex]
a. P(X<30)
This is the pvalue of Z when X = 30. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30 - 33}{1.7}[/tex]
[tex]Z = -1.76[/tex]
[tex]Z = -1.76[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{32 - 33}{1.7}[/tex]
[tex]Z = -0.59[/tex]
[tex]Z = -0.59[/tex] has a pvalue of 0.2776.
X = 28
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{28 - 33}{1.7}[/tex]
[tex]Z = -2.94[/tex]
[tex]Z = -2.94[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{35 - 33}{1.7}[/tex]
[tex]Z = 1.18[/tex]
[tex]Z = 1.18[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{31 - 33}{1.7}[/tex]
[tex]Z = -1.18[/tex]
[tex]Z = -1.18[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 33}{1.7}[/tex]
[tex]X - 33 = 1.645*1.7[/tex]
[tex]X = 35.7965[/tex]
At least 35.7965 mpg
The upper 5% of cars have a mileage rating of 35.805 mpg
What is z score?
Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean) / standard deviation
Given; mean of 33 mpg and a standard deviation of 1.7
a) For < 30:
z = (30 - 33)/1.7 = -1.76
P(x < 30) = P(z < -1.76) = 1 - 0.8413 = 0.0392
b) For < 28:
z = (28 - 33)/1.7 = -2.94
P(x < 28) = P(z < -2.94) = 0.0016
c) For > 35:
z = (35 - 33)/1.7 = 1.18
P(x > 35) = P(z > 1.18) = 1 - P(z < 1.18) = 1 - 0.8810 = 0.119
d) For > 31:
z = (31 - 33)/1.7 = -1.18
P(x > 31) = P(z > -1.18) = 1 - P(z < -1.18) = 0.8810
e) The upper 5% of cars achieve have a z score of 1.65, hence:
1.65 = (x - 33)/1.7
x = 35.805 mpg
The upper 5% of cars have a mileage rating of 35.805 mpg
Find out more on z score at: https://brainly.com/question/25638875
