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.

 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

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

The standardized score with 5% below is -1.6449; $-1.6449=\frac{x-39}{6}⇒x=\left(-1.6449\right)\left(6\right)+39\approx 29.1309$. 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.

 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 $\stackrel{^}{p}$ be the proportion of applicants at this location that will be accepted.

 11 Find ${\mu }_{\stackrel{^}{p}}$. (2)

${\mu }_{\stackrel{^}{p}}=p=0.7$

 12 Find ${\sigma }_{\stackrel{^}{p}}$. (5)

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

 13 Describe the shape of the sampling distribution of $\stackrel{^}{p}$. 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)

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

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

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

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