AP Statistics |
Unit Test #8 |

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The time required for an office to process paperwork varies approximately normally with mean 39 days and standard deviation 6 days.

Use this information for all questions on this page.

1. |
Calculate and interpret the standardized score of a piece of paperwork that takes 50 days to process. |
(5) |

$\frac{50-39}{6}\approx 1.8333$; this is above average, but not unusually so.

2. |
According to the Empirical Rule, what approximate percentage of papers will be processed in 27 days or less? |
(3) |

Since 27 days is 2 standard deviations below the mean, about 2.5% of papers will take less than 27 days.

3. |
According to the Empirical Rule, how long does it take to process the slowest 16% of papers? |
(3) |

About 16% of the distribution lies one standard deviation above the mean; $39+6=45$, so the slowest 16% take 45 days or longer.

4. |
Find the probability that a randomly selected paper will require between 30 and 40 days. |
(5) |

$P\left(30<X<40\right)=P\left(\frac{30-39}{6}<z<\frac{40-39}{6}\right)=P\left(-1.5<z<0.1667\right)\approx 0.49938$

5. |
The 5% of papers with the fastest processing times require how much time? |
(5) |

The standardized score with 5% below is -1.6449; $-1.6449=\frac{x-39}{6}\Rightarrow x=\left(-1.6449\right)\left(6\right)+39\approx 29.1309$. The fastest 5% of papers take 29.1309 days or less.

Census records show that the amount of time that adults spend watching television (*X*)
varies approximately normally with mean 5 hours and standard deviation 1.3 hours. A group of ten adults are
randomly selected, and their television habits are recorded.

Use this information for all questions on this page.

6. |
Find ${\mu}_{\overline{x}}$. |
(2) |

${\mu}_{\overline{x}}=\mu =5$

7. |
Find ${\sigma}_{\overline{x}}$. |
(4) |

${\sigma}_{\overline{x}}=\frac{\sigma}{\sqrt{n}}=\frac{1.3}{\sqrt{10}}\approx 0.4111$

8. |
Describe the shape of the sampling distribution of $\overline{x}$. Justify your answer. |
(4) |

The shape will be approximately normal, because the population is approximately normal.

9. |
Find the probability that the mean weight time spent watching television by this group was greater than 5.5 hours. |
(5) |

$P\left(\overline{x}>5.5\right)=P\left(z>\frac{5.5-5}{0.4111}\right)=P\left(z>1.2163\right)\approx 0.1119$

10. |
Find the probability that the mean weight time spent watching television by this group was less than four hours. |
(5) |

$P\left(\overline{x}<4\right)=P\left(z<\frac{4-5}{0.4111}\right)=P\left(z<-2.4325\right)\approx 0.0075$

A large job training program has historically accepted 70% of the applicants to the program. One office for this program had 40 applicants in a recent year. Let $\widehat{p}$ be the proportion of applicants at this location that will be accepted.

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11. |
Find ${\mu}_{\widehat{p}}$. |
(2) |

${\mu}_{\widehat{p}}=p=0.7$

12. |
Find ${\sigma}_{\widehat{p}}$. |
(5) |

${\sigma}_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}=\sqrt{\frac{\left(0.7\right)\left(0.3\right)}{40}}\approx 0.0725$

13. |
Describe the shape of the sampling distribution of $\widehat{p}$. Justify your answer. |
(5) |

The shape will be approximately normal because $np=\left(40\right)\left(0.7\right)=28>10$ and $n\left(1-p\right)=\left(40\right)\left(0.3\right)=12>10$.

14. |
Find the probability that at least 32 of the applicants will be accepted. |
(5) |

$P\left(X\ge 32\right)=P\left(\widehat{p}>\frac{32}{40}\right)=P\left(\widehat{p}>0.8\right)=P\left(z>\frac{0.8-0.7}{0.0725}\right)=P\left(z>1.3801\right)\approx 0.0838$

15. |
Find the probability that fewer than 25 of the trainees will be accepted. |
(5) |

$P\left(X\le 25\right)=P\left(\widehat{p}<\frac{25}{40}\right)=P\left(\widehat{p}<0.625\right)=P\left(z<\frac{0.625-0.7}{0.0725}\right)=P\left(z<-1.0351\right)\approx 0.1503$

Page last updated 10:26 2020-01-30