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 temperature of a single cup is between 158°F and 162°F?
- The probability is found by finding the area between the two values on a standardized normal distribution using a Z-Table or a calculator
- On a Z-Table,
- The z-score of 158$\degree$ is $z = \frac{158 - 160}{5} = \frac{-2}{5} = -0.40
- The z-score of 162$\degree$ is $z = \frac{162 - 160}{5} = \frac{2}{5} = 0.40
- Look up area left of z = -0.60
- Look up area left of z = 0.60
- Take the difference between the two
- $P(-0.40 < z < 0.40) = P(z < 0.40) - P(z < -0.40) = 0.6554 - 0.3446 = 0.311$
- On a TI-83/84, DISTR –> 2:normalcdf(
- 2:normalcdf(158,162,160,5) if using the values from the problem
- 2:normalcdf(0.3446,0.6554,0,1) if using the z-score (gives the same answer)
- Probability = 0.311 = 31.1%
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