5.8 Review Questions
- The standard normal distribution has mean ________ and standard deviation ________.
- The area under the density curve that lies between 30 and 40 is 0.832. What percentage of all possible observations of the variable are either less than 30 or greater than 40?
- The finishing times in the New York City 10-km run are normally distributed with mean 61 minutes and standard deviation 9 minutes.
- Determine the percentage of finishers with times between 50 and 70 minutes.
- Determine the percentage of finishers with times less than 75 minutes.
- Obtain and interpret the 40th percentile for the finishing times.
- Find and interpret the 8th decile for the finishing times.
Show/Hide Answer
- The standard normal distribution has a mean 0 and
standard deviation 1. - Since the total area under any density curve is 1, the total area to the left of 30 and to the right of 40 is 1-0.832=0.168. The percentage of observations that are either less than 30 or greater than 40 is 16.8%.
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- Want to find the area between 50 and 70, we need to find the z-scores first and then use Table II.
[latex]\begin{align*}P(50 < X < 70) &= P(\frac{50 - 61}{9} < Z < \frac{70 - 61}{9}) = P( - 1.22 < Z < 1) \\ &= P(Z< 1) - P(Z < - 1.22) = 0.8413 - 0.1112 = 0.7301. \end{align*}[/latex]
There are 73.01% of finishers with times between 50 and 70 minutes. We could also double check with R commander, the results are close. The results differ since we round the z score to two decimals and use the table.
## [1] 0.8413447 0.1108118
## [1] 0.7305329
- If you want to find the area to the left of 75, find the z-score first and then use Table II to find the area. \[P(X < 75) = P(Z < \frac{75 – 61}{9}) = P(Z < 1.56) = 0.9406.\] There are 94.06% of finishers with times less than 75 minutes. We could also double check with R commander, the results are close.
## [1] 0.9400931
- Want to know 40% of finishers with times below what value. First, use Table II to find the z-score with the area to its left is 0.4. The closest value in the main body of Table II is 0.4013 with the corresponding z-score of -0.25. Therefore, the normal quantile is
[latex]x = \mu + z \times \sigma = 61 + ( - 0.25) \times 9 = 58.75.[/latex]
Interpretation: There are 40% of finishers with times less than 58.75 minutes.
We could also double check with R commander, the results are close.
## [1] 58.71988
- 8th decile is the same as the 80th percentile. Want to know 80% of finishers with times below what value. First, use Table II to find the z-score with the area to its left is 0.8. The closest value in the main body of Table II is 0.7995, with the corresponding z-score of 0.84.
Therefore, the normal quantile is [latex]x = \mu + z \times \sigma = 61 + 0.84 \times 9 = 68.56.[/latex]
Interpretation: 80% of finishers have times less than 68.56 minutes.
We could also double check with R commander, the results are close.
## [1] 68.57459
- Want to find the area between 50 and 70, we need to find the z-scores first and then use Table II.
