AP Statistics |
Unit Test #8 |

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The fuel economy for a car varies approximately normally with mean 24.8 mpg and standard deviation 6.2 mpg. Use this information for all questions on this page.

1. |
Calculate and interpret the standardized score of a car that has a fuel economy of 19 mpg. |
(5) |

$\frac{19-24.8}{6.2}=-0.9355$; this car’s fuel economy is 0.9355 standard deviations below the mean (which is not unusual).

2. |
According to the Empirical Rule, what approximate percentage of cars have fuel economies of 18.6 mpg or less? |
(3) |

Since 18.6 is one standard deviation below the mean, about 16% of cars will have fuel economies of 18.6 mpg or less.

3. |
According to the Empirical Rule, approximately 2.5% of cars have fuel economies that are greater than what amount? |
(3) |

The upper 2.5% of fuel economies will have standardized scores of 2 or more; that translates to fuel economies of $\left(2\right)\left(6.2\right)+24.8=37.2$ mpg or more.

4. |
Find the probability that a randomly selected car will have a fuel economy between 18 mpg and 25 mpg. |
(5) |

$P\left(18<x<25\right)=P\left(\frac{18-24.8}{6.2}<z<\frac{25-24.8}{6.2}\right)=P\left(-1.0968<z<0.0323\right)\approx 0.3765$

5. |
The 1% of cars with the worst fuel economies get how many miles to the gallon? |
(5) |

The lowest 1% will have z-scores of $-2.3263$ and lower. That translates to fuel economies of $x=\left(-2.3263\right)\left(6.2\right)+24.8=10.3766$ mpg or lower.

An airline has found that weight of a piece of luggage brought by a passenger (*X*) varies
with mean 40 pounds and standard deviation 8 pounds. A particular plane can carry 200 passengers—assume
that each passenger brings one piece of luggage.

Use this information for all questions on this page.

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

${\mu}_{\overline{x}}=\mu =40$

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

${\sigma}_{\overline{x}}=\frac{\sigma}{\sqrt{n}}=\frac{8}{\sqrt{200}}=0.5657$

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

Since the size of the sample is quite large (well over 30), the shape of the sampling distribution of $\overline{x}$ will be approximately normal.

9. |
Find the probability that the mean weight of the luggage is greater than 41 pounds. |
(5) |

$P\left(\overline{x}>41\right)=P\left(z>\frac{41-40}{0.5657}\right)=P\left(z>1.7678\right)\approx 0.0386$

10. |
Find the probability that the mean weight of the luggage is less than 39.5 pounds. |
(5) |

$P\left(\overline{x}<39.5\right)=P\left(z<\frac{39.5-40}{0.5657}\right)=P\left(z<-0.8839\right)\approx 0.1884$

A large corporation has found, over the years, that about 10% of its sales trainees are rated as outstanding at the completion of their training program. A particular office of this corporation had 150 sales trainees in a recent year. Let $\widehat{p}$ be the proportion of sales trainees at this location that will be rated as outstanding.

Use this information for all questions on this page.

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

${\mu}_{\widehat{p}}=p=0.1$

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

${\sigma}_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}=\sqrt{\frac{\left(0.1\right)\left(0.9\right)}{150}}\approx 0.0245$

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

The shape will be approximately normal because $np=\left(150\right)\left(0.1\right)=15>10$ and $n\left(1-p\right)=\left(150\right)\left(0.9\right)=135>10$.

14. |
Find the probability that at least 17 of the trainees will be rated as outstanding. |
(5) |

$P\left(X\ge 17\right)=P\left(\widehat{p}>\frac{17}{150}\right)=P\left(\widehat{p}>0.1133\right)=P\left(z>\frac{0.1133-0.1}{0.0245}\right)=P\left(z>0.5443\right)\approx 0.2931$

15. |
Find the probability that fewer than 14 of the trainees will be rated as outstanding. |
(5) |

$P\left(X<14\right)=P\left(\widehat{p}<\frac{14}{150}\right)=P\left(\widehat{p}<0.0933\right)=P\left(z<\frac{0.0933-0.1}{0.0245}\right)=P\left(z<-0.2722\right)\approx 0.3927$

Page last updated 10:26 2020-01-30