5. Probabilit omputations using the standard normal distribution #2 Aa Aa An aut

Question

5. Probabilit omputations using the standard normal distribution #2 Aa Aa An automobile bo ery manufacturer offers a 22/37 warranty on its batterles. The first number in the warranty code is the free-replacement perlod; the second number is the prorated-credit period. Under this warranty, if a battery fails within 22 months of purchase, the manufacturer replaces the battery at no charge to the consumer. If the battery fails after 22 months but within 37 months, the manufacturer provides a prarated credit toward the purchase of a new battery. The manufacturer assumes that X, the lifetime of its auto batteries, is normally distributed with a mean af 30 marths and a standard deviation of 3.5 months. Use the fallawing Distributions tool to help you answer the questions that follow. (Hint: When you adjust the parameters of a distribution, you must reposition the vertical line (or lines) for the correct areas to be displayed.) Norma Mean30 Standard Deviation5.5 51015 20 25203543-45 5# 55 of its batteries free of charge If the manufacturers assumptions are correct, it would need to replace he company finds that it is replaong 1.54% of its batteries free of charge. It suspects that rs assumption about the standard deviation of the life of its batteries is incorrect. A standard deviation of replacement rate results in a 154% using the revised standard deviation for battery life, what percentage of the free replacement but do qualify for the prorated credit? O 48-46 95.52% O 47.06 o 2.94%

Explanation / Answer

Solution:- mean = 30 and standard deviation = 3.5

1) P(X<22) = P(Z < (22-30)/3.5) = P(Z < -2.2857) = 1 0.989 =0.011

=> if the manufacture's assumptions are correct.it would need to replace 1.1% of the batteries free of charge.

2)

Z-value of 0.0154 is -2.1597

using central limit theorem
- 2.1597 = (22 - 30)/sigma
sigma = 3.7042

=> A standard deviation of 3.7042 results is a 1.54% replacement rate.

3) option B. 95.52%

=> P(22 < X < 37) = P((22 - 30)/3.7042 < z < (37 - 30)/3.7042) = P(-2.1597 < z < 1.8897) = 0.9552

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