What percentage of data would you predict would be between 40 and 70 and what percentage would you predict would be more than 70
2 answers:
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Answer:
a) 0.1728
b) 0.09183
c) 0.7354
d) $ 222.25
Explanation:
Given
mean = = $367
Standard deviation = =$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, ,
, 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, ,
)
a = NORMINV ( 0.05 , 367, 88 ) = 222.2529
Cost = $ 222.25
Answer:
Machine B EAC is $17,705.78 more than Machine A EAC.
Explanation:
First find the present values of the cost of both machines.
Machine A:
= 200,000 + (15,000 * Present value of annuity interest factor, 15%, 8 years)
= 200,000 + ( 15,000 * 4.4873)
= $267,309.50
Machine B
= 300,000 + (17,500 * Present value of annuity interest factor, 15%, 10 years)
= 300,000 + 17,500 * 5.0188
= $387,829
Equivalent Annual cost Machine A:
= [(NPV * Required return) / 1 - (1 + Required return) ^–Number of Periods
=[(267,309.50 * 15%) / 1 - 1.15⁻⁸
= $59,569.95
Equivalent Annual cost Machine B:
= (387,829 * 15%) / (1 - 1.15⁻¹⁰)
= $77,275.73
Difference:
= 77,275.73 - 59,569.95
= $17,705.78
