Answer to Practice Problem 3
A coffee shop claims that the average temperature of its freshly brewed coffee is 160°F with a standard deviation of 5°F. A health inspector randomly samples 41 cups of coffee.
- What is the probability that the sample mean temperature is between 158°F and 162°F?
- The mean and standard deviation are:
- $\mu_{\bar{x}}$ = $\mu$ = 160
- $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{41}} = \frac{5}{6.403} = 0.7809$
- 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 of 158$\degree$ is $z = \frac{158 - 160}{0.7809} = \frac{-2}{0.7809} = -2.5612
- The z-score of 162$\degree$ is $z = \frac{162 - 160}{0.7809} = \frac{2}{0.7809} = 2.5612
- Look up area left of z = -2.5612
- $P(z < -2.5612) = 0.00522$
- Look up area left of z = 0.60
- $P(z < 2.5612) = 0.99478$
- Take the difference between the two
- $P(-2.5612 < z < 2.5612) = P(z < 2.5612) - P(z < -2.5612) = 0.99478 - 0.00522 = 0.9896$
- On a TI-83/84, DISTR –> 2:normalcdf(
- 2:normalcdf(158,162,160,0.7809) if using the values from the problem
- 2:normalcdf(-2.5612,2.5612,0,1) if using the z-score (gives the same answer)
- Probability = 0.9896 = 98.96%
Click here to return to the lecture notes.