Question

The length of time a 1999 Honda Civic lasts before needing major repairs is normally distributed with a mean of 7 years and a standard deviation of 1.7 years.

a. What is the probability that a randomly selected 1999 Honda Civic lasts between 5.5 and 9 years?

b. If Honda Motor Company, Ltd. informed you that your Honda Civic is at 70th percentile, what is your Honda Civic’s years without major repairs?

c. A family-owned two Honda Civics. One of them is 1999 and lasted 7.5 years without a major repair. Other one (2002) lasted 9.0 years, but the distribution of 2002 Honda Civic had higher mean (8.4) with a standard deviation of 1.9 years. Which Honda Civic did better relative to other Honda Civics from the same year? (hint: Need to compare z scores from different normal distributions (1999 and 2002).

Answer 1

Given = 7 and = 1.7

To find the probability, we need to find the z scores.

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(a) For P (5.5 < X < 9) = P(X < 9) – P(X < 5.5)

For P( X < 9)

Z = (9 – 7)/1.7 = 1.18

The probability for P(X < 9) from the normal distribution tables is = 0.8810

For P( X < 5.5)

Z = (5.5 – 7)/1.7 = -0.88

The probability for P(X < 5.5) from the normal distribution tables is = 0.1894

Therefore the required probability is 0.8810 – 0.1894 = 0.6916

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(b) 70th percentile = 70% = 0.7 i.e P(X < x) = 0.7

From the tables, the z value =

0.5244 = (X - 7) / 1.7

Solving for X, we get X = (0.5244 * 1.7) + 7 = 7.89 years

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(c) 1999 lasted 7.5 years

Z = (7.5 - 7) / 1.7 = 0.29

2002 lasted 9 years. Given = 8.4 and = 1.9

Z = (9 - 8.4) / 1.9 = 0.32

Therefore the 2002 Honda Civic did better than the 1999 Honda Civic

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