Q&A: The lifetime of a 2-volt non-rechargeable battery in constant use has...

...a mean of 516 hours and a standard deviation of 60 hours. A box of 100 of these batteries is considered a random sample, what is the probability the average lifetime of the batteries in the box are shorter than 505 hours?

A: make use of the central limit theorem.

  • compute sample standard deviation: because the sample size is > 30, the distribution of the x-bars is approximately normal with mean 516 and stanadrd deviation s/N^0.5 = 60/sqrt(100) = 6
  • compute the z-score for 500: (505 - 516)/6 = -1.83
  • P(Z<-1.83) = 3.3% = P(X-bar < 505)

Bonus Points

  • The central limit theorem states that given a distribution with a mean m and variance s^2, the sampling distribution of the mean approaches a normal distribution with a mean (m) and a variance s^2/N as N increases.
    • N, the sample size, in general has to be > 30.
    • N is the sample size for each mean and not the number of samples taken.
    • As N grows bigger, the distributions become more and more normal, the spread of the distributions decreases.
  • the theorem can be applied on a population with any probability distribution.

Category: Quantitative Analysis > Probability

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