question archive A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features
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A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 47 months and a standard deviation of 9 months. Using the 68-95-99.7 rule, what is the approximate percentage of cars that remain in service between 56 and 65 months?
Answer:
Step-by-step explanation
The 68-95-99.7 rule defined when X is an observation from a random bell-shaped (normally distributed) value with mean (µ) and standard deviation (σ), having these following probabilities:
Pr(µ - σ ≤ X ≤ µ + σ)
Pr(µ - 2σ ≤ X ≤ µ + 2σ)
Pr(µ - 3σ ≤ X ≤ µ + 3σ)
Given: mean = 47 months and SD = 9 months. Substitute:
Pr(47 - 9 ≤ X ≤ 47 + 9) = Pr( 38 ≤ X ≤ 56) = 0.6826
Pr(47 - 18 ≤ X ≤ 47 + 18) = Pr(29 ≤ X 65) = 0.9544
Pr(47 - 27 ≤ X ≤ 47 + 27) = Pr(20 ≤ X ≤ 74) = 0.9974
Between 56 and 65 minutes is between one and two standard deviations above the mean.
Pr(47 - 9 ≤ X ≤ 47 + 9) = 0.6826 and Pr(47 - 18 ≤ X ≤ 47 + 18) = 0.9544
Find the percentage above the mean, by subtracting this two values:
P (0.9544 - 0.6826) = 0.2718 or 27.18%
The approximate percentage of cars that remain in service between 56 and 65 months is 27.18%.