AP Stats - Test #8

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.

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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.

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.

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.

Find the probability that a randomly selected paper will require between 30 and 40 days.

(5)

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

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

(5)

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

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Find $μ_{\bar{x}}$ .

(2)

$μ_{\bar{x}} = μ = 5$

Find $σ_{\bar{x}}$ .

(4)

$σ_{\bar{x}} = \frac{σ}{\sqrt{n}} = \frac{1.3}{\sqrt{10}} \approx 0.4111$

Describe the shape of the sampling distribution of $\bar{x}$ . Justify your answer.

(4)

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

Find the probability that the mean weight time spent watching television by this group was greater than 5.5 hours.

(5)

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

10.

Find the probability that the mean weight time spent watching television by this group was less than four hours.

(5)

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

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

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11.

Find $μ_{\hat{p}}$ .

(2)

$μ_{\hat{p}} = p = 0.7$

12.

Find $σ_{\hat{p}}$ .

(5)

$σ_{\hat{p}} = \sqrt{\frac{p (1 - p)}{n}} = \sqrt{\frac{(0.7) (0.3)}{40}} \approx 0.0725$

13.

Describe the shape of the sampling distribution of $\hat{p}$ . Justify your answer.

(5)

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

14.

Find the probability that at least 32 of the applicants will be accepted.

(5)

$P (X \geq 32) = P (\hat{p} > \frac{32}{40}) = P (\hat{p} > 0.8) = P (z > \frac{0.8 - 0.7}{0.0725}) = P (z > 1.3801) \approx 0.0838$

15.

Find the probability that fewer than 25 of the trainees will be accepted.

(5)

$P (X \leq 25) = P (\hat{p} < \frac{25}{40}) = P (\hat{p} < 0.625) = P (z < \frac{0.625 - 0.7}{0.0725}) = P (z < - 1.0351) \approx 0.1503$

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