Respuesta :
Answer:
[tex]\mu = 367[/tex]
[tex]\sigma = 88[/tex]
a. What is the probability that the cost will be more than $450?
We are supposed to find P(x > 450)
Formula : [tex]z=\frac{x-\mu}{\sigma}[/tex]
[tex]z=\frac{450-367}{88}[/tex]
[tex]z=0.9431[/tex]
Refer the z table :
P(z<0.9431)=0.8264
P(z> 0.9431)=1-P(z<0.9431)=1- 0.8264= 0.1736
Hence the probability that the cost will be more than $450 is 0.1736
b)What is the probability that the cost will be less than $250?
We are supposed to find P(x<250)
Formula : [tex]z=\frac{x-\mu}{\sigma}[/tex]
[tex]z=\frac{250-367}{88}[/tex]
[tex]z=-1.3295[/tex]
Refer the z table :
P(z<-1.3295)=0.0934
Hence the probability that the cost will be less than $250 is 0.0934
c)What is the probability that the cost will be between $250 and $450?
P(250<z<450)=P(z<450)-P(z<250) = 0.8264 - 0.0934 =0.733
Hence the probability that the cost will be between $250 and $450 is 0.733
d) If the cost for your car repair is in the lower 5% of automobile repair charges, what is your cost?
p = 0.05
refer the z table
z = -1.65
Formula : [tex]z=\frac{x-\mu}{\sigma}[/tex]
[tex]-1.65=\frac{x-367}{88}[/tex]
[tex]-1.65 \times 88=x-367[/tex]
[tex]-145.2+367=x[/tex]
[tex]221.8=x[/tex]
Hence If the cost for your car repair is in the lower 5% of automobile repair charges, so, cost is $221.8
Using the normal distribution, we have that:
a) There is a 0.8264 - 82.64% probability that the cost will be more than $450.
b) There is a 0.0918 = 9.18% probability that the cost will be less than $250.
c) There is a 0.7346 = 73.46% probability that the cost will be between $250 and $450.
d) The cost will be of at most $222.
In a normal distribution 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]
It 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, which is the percentile of X.
In this problem:
- Mean of $367, thus [tex]\mu = 367[/tex]
- Standard deviation of $88, thus [tex]\sigma = 88[/tex].
Item a:
This probability is 1 subtracted by the p-value of Z when X = 450, so:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{450 - 367}{88}[/tex]
[tex]Z = 0.94[/tex]
[tex]Z = 0.94[/tex] has a p-value of 0.8264.
0.8264 - 82.64% probability that the cost will be more than $450.
Item b:
This probability is the p-value of Z when X = 250, so:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{250 - 367}{88}[/tex]
[tex]Z = -1.33[/tex]
[tex]Z = -1.33[/tex] has a p-value of 0.0918.
0.0918 = 9.18% probability that the cost will be less than $250.
Item c:
This probability is the p-value of Z when X = 450 subtracted by the p-value of Z when X = 250, so:
0.8264 - 0.0918 = 0.7346.
0.7346 = 73.46% probability that the cost will be between $250 and $450.
Item d:
The cost is of at most the 5th percentile, which is X when Z has a p-value of 0.05, so X when Z = -1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.645 = \frac{X - 367}{88}[/tex]
[tex]X - 367 = -1.645(88)[/tex]
[tex]X = 222[/tex]
The cost will be of at most $222.
A similar problem is given at https://brainly.com/question/13847266
