Answer to Practice Problem 2
A factory produces light bulbs with a mean lifetime of 1,200 hours and a standard deviation of 100 hours. A quality control engineer selects a random sample of 36 bulbs.
- What is the probability that the sample mean lifetime of the 36 bulbs is greater than 1,225 hours?
- The mean and standard deviation are:
- $\mu_{\bar{x}}$ = $\mu$ = 1200
- $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{100}{\sqrt{36}} = \frac{100}{6} = 16.667$
- The probability is found by finding the area of the right tail of the sampling distribution using a Z-Table or a calculator
- On a Z-Table,
- The z-score is $z = \frac{1225 - 1200}{16.667} = \frac{25}{16.667} = 1.5$
- Look at the area left of z = 1.5
- Take the compliment to get the area to the right of z = 0.25
- $P(z > 1.5) = 1 - P(z < 1.5) = 1 - 0.933 = 0.067$
- On a TI-83/84, DISTR –> 2:normalcdf(
- 2:normalcdf(1225,99999,1200,16.667) if using the values from the problem
- 2:normalcdf(1.5,9999,0,1) if using the z-score (gives the same answer)
- Probability = 0.067 = 6.7%
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