Automobile repair costs continue to rise with the average cost now at $367 per repair.† Assume that the cost for an automobile repair is normally distributed with a standard deviation of $88.
A. What is the probability that the cost will be more than $450?
B. What is the probability that the cost will be less than $250?
C. What is the probability that the cost will be between $250 and $450?
D. If the cost for your car repair is in the lower 5% of automobile repair charges, what is your cost?

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funmilaciousfunmie78 funmilaciousfunmie78 · Answer 1 · 2020-07-01T02:12:22+02:00

Answer:

a) 0.1728

b) 0.09183

c) 0.7354

d) $ 222.25

Explanation:

Given

mean = [tex]\mu[/tex] = $367

Standard deviation = [tex]\sigma[/tex] =$88

Cost of automobile repair is normally distributed.

a) We have to find P( x > 450 )

P( x > 450 ) = 1 - P( x <= 450 )

Using excel function, P( x <= x ) = NORMDIST (x, [tex]\mu[/tex], [tex]\sigma[/tex], 1 )

P( x > 450 ) = 1 - NORMDIST( 450 , 367, 88, 1 )

= 1 - 0.8272 = 0.1728

P( x > 450 ) = 0.1728

b) P( x < 250 ) = NORMDIST( 250 , 367, 88, 1 ) = 0.09183

P( x < 250 ) = 0.09183

c) P( 250 < x < 450 ) = P( x <450 ) - P( x < 250 )

P( x <450 ) = NORMDIST( 450 , 367, 88, 1 ) = 0.8272

P( x < 250 ) = NORMDIST( 250 , 367, 88, 1 ) = 0.09183

P( 250 < x < 450 ) = 0.8272 - 0.09183 = 0.7354

P( 250 < x < 450 ) = 0.7354

d) We have P( X < a ) = 0.05

We have to find a.

Using Excel, = NORMINV ( Probability, [tex]\mu[/tex], [tex]\sigma[/tex] )

a = NORMINV ( 0.05 , 367, 88 ) = 222.2529

Cost = $ 222.25