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Probability in Normal Distribution

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1. A certain brand of electric stove has a length of life that is approximately normally distributed with a mean life of 15.7 years and a standard deviation of 4.2 years (Blaisdell, 1998).

a. What percentage of stoves would we expect to last less than 13.5 years?

b. What percentage of stoves would we expect to last between 10 and 20 years?

c. These stoves are covered by a warrantee for five years. What percentage of stoves would we expect to have to be replaced during the warrantee period?

d. What should the warrantee period be if the company wants to make sure that no more than 2% of the stoves would have to be replaced? Explain and show your work.

2. Suppose that 10 ft lengths of a certain type of cable have breaking strengths that are normally distributed with a mean of 450 lbs and a standard deviation of 50 lbs. If five of these cables are intertwined and used to support a single total load L, suppose that the cables are loaded equally in such a way that the support will fail if any one of the cables fails (that is, if it has a breaking strength below L).

a. Suppose that the total load we need our 5 cables to hold is 2000 pounds. What is the probability that a single cable will fail individually?

b. Find the probability that the 5 cable support will fail. Please be sure to use proper notation and to show all of your work.

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Solution Summary

The probability in normal distributions are examined. The percentages of stoves which would expected to last less 13.5 years is determined.

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1a.
Let X be the life of an electric stove. Then X follows normal distribution with mean 15.7 and standard deviation 4.2. Set Z=(X-15.6)/4.2. Then Z ~ N(0, 1). The percentage of stoves which we would expect to last less than 13.5 years is
P(X<13.5)= P(Z<-0.5)=0.3085 =30.85%

b.
Let X be the life of an electric stove. Then X follows normal distribution with mean 15.7 and standard deviation 4.2. Set Z=(X-15.6)/4.2. Then Z ~ N(0, 1). The percentage of stoves which we would ...

Solution provided by:
Changping Wang, MA
Education
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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