Assume the cost of an extended 100,000 mile warranty for a particular SUV follow

Question

Assume the cost of an extended 100,000 mile warranty for a particular SUV follows the normal distribution with a mean of $1.620 and a standard deviation of $85. Complete parts (a) through (d) below a) Determine the interval of warranty costs from various companies that are one standard deviation around the mean The interval of warranty costs that are one standard deviation around the mean ranges from S toS (Type integers or decimals. Use ascending order.) b) Determine the interval of warranty costs from various companies that are two standard deviations around the mean The interval of warranty costs that are two standard deviations around the mean ranges from S toS (Type integers or decimals. Use ascending order.)

Explanation / Answer

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 1620
standard Deviation ( sd )= 85
A.
68% OF DATA
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= [1620 ± 85]
= [ 1620 - 85 , 1620 + 85]
= [ 1535 , 1705 ]
B.
95% OF DATA
About 95% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 2s.d)
So to the given normal distribution about 95% of the observations lie in between
= [1620 ± 2 * 85]
= [ 1620 - 2 * 85 , 1620 + 2* 85]
= [ 1450 , 1790 ]
C.
99.7% OF DATA
About 99.7% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 3s.d)
So to the given normal distribution about 99.7% of the observations lie in between
= [1620 ± 3 * 85]
= [ 1620 - 3 * 85 , 1620 + 3* 85]
= [ 1365 , 1875 ]
D.
The $1960 cost of this warranty is much higher than average due
to the fact that it is more than three standard deviations above the
m9an.

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