Chapter 7 Extra Extra Practice Name______

AP Stats

13. The distribution of flood policies sold is approximately normal. The amount of flood insurance policies sold in the last four days by a new agent were recorded as 3, 5, 2, and 7. Assume that samples of size 2 are randomly selected with replacement from this population of four values.

a) List the different possible samples of size 2. {1 point}

b) Find the mean of each sample. {3 points}

c) Find the mean of the sampling distribution, in other words what is the mean of these means? c) ____4.25___

Round to two decimal places. {1 point}

d) Determine the mean of the population. {1 point} d) ____4.25____

e) Is mean a biased or unbiased estimator? {1 point} e) ____yes_____

2. Suppose that the number of movies viewed in the last year by high school students has an average of 19.3 with a standard deviation of 15.8. Suppose we take an SRS of 100 high school students and calculate the mean number of movies viewed by the members of the sample.

a) What is the mean of the sampling distribution of?

b) What is the standard deviation of the sampling distribution of? Check whether the 10% condition is satisfied.

. The 10% condition is met because there are more than 10(100) = 1000 high school students.

3. The superintendent of a large school district wants to know what proportion of middle school students in her district are planning to attend a four-year college or university. Suppose that 80% of all middle school students in her district are planning to attend a four-year college or university. What is the probability that a SRS of size 125 will give a result within 7 percentage points of the true value?

State: We want to find the probability that the proportion of middle school students who plan to attend a four-year college or university falls between 73% and 87%. That is, P(0.73 £ £ 0.87).

. Because the school district is large, we can assume that there are more than 10(125) = 1250 middle school students so. We can consider the distribution of to be approximately Normal since np = 125(0.80) = 100 ³ 10 and n(1 – p) = 125(0.20) = 25 ³ 10.

.692 .728 .764 .8 .836 .872 .908

Calculations: P(0.73 £ £ 0.87) = normalcdf(lower: 0.73, upper: 0.87, µ: 0.80, σ:.036) = 0.9495

Answer: About 95% of all SRSs of size 125 will give a sample proportion within 7 percentage points of the true proportion of middle school students who are planning to attend a four-year college or university.

4. Suppose that the number of texts sent during a typical day by a randomly selected high school student follows a right-skewed distribution with a mean of 15 and a standard deviation of 35. Assuming that students at your school are typical texters, how likely is it that a random sample of 50 students will have sent more than a total of 1000 texts in the last 24 hours?

State: What is the probability that the total number of texts in the last 24 hours is greater than 1000 for a random sample of 50 high school students?

A total of 1000 texts among 50 students is the same as an average number of texts of. We want to find P(> 20), where

= sample mean number of texts.

Since n is large (50 > 30), is approximately Normal

.15 5.1 10.05 15 19.95 24.9 29.85

Calculater: P(> 20) ≈ normalcdf = 0.1562.

Conclude: There is about a 16% chance that a random sample of 50 high school students will send more than 1000 texts in a day.

5. At the P. Nutty Peanut Company, dry roasted, shelled peanuts are placed in jars by a machine. The distribution of weights in the bottles is approximately Normal, with a mean of 16.1 ounces and a standard deviation of 0.15 ounces.

a) Without doing any calculations, explain which outcome is more likely, randomly selecting a single jar and finding the contents to weigh less than 16 ounces or randomly selecting 10 jars and finding the average contents to weigh less than 16 ounces.

Since averages are less variable than individual measurements, I would expect the sample mean of 10 jars to be closer, on average, to the true mean of 16.1 ounces. Thus, it is more likely that a single jar would weigh less than 16 ounces than the average of 10 jars to be less than 16 ounces.

b) Find the probability of each event described above.

Let x = weight of the contents of a randomly selected jar of peanuts. x is N(16.1, 0.15).

P(x < 16) = normalcdf(lower: –¥, upper: 16, µ: 16.1, s: 0.15) = 0.2525.

Let= average weight of the contents of a random sample of 10 jars. is.

DRAW A PICTURE AND SHADE!!!!

P(< 16) = normalcdf= 0.0175. This answer agrees with the answer to part (a) because this probability is much smaller than 0.2525.