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**Correct Answer: A AP1.14. Boxplots of two data sets are shown.**

Based on the boxplots, which statement below is true? (a) The spread of both plots is about the same. (b) The means of both plots are approximately equal. (c) Plot 2 contains more data points than Plot 1. (d) The medians are approximately equal. (e) Plot 1 is more symmetric than Plot 2. AP1.14. Boxplots of two data sets are shown. Based on the boxplots, which statement below is true? (a) The spread of both plots is about the same. (b) The means of both plots are approximately equal. (c) Plot 2 contains more data points than Plot 1. (d) The medians are approximately equal. (e) Plot 1 is more symmetric than Plot 2. Correct Answer: A

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**The Sampling Distribution of : Mean and Standard Deviation**

Suppose that is the mean of an SRS of size n drawn from a large population with mean μ and standard deviation σ. Then:

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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 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. 1.58

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**DEFINITION: Central limit theorem (CLT)**

Draw an SRS of size n from any population with mean μ and finite standard deviation σ. The central limit theorem (CLT) says that when n is large, the sampling distribution of the sample mean is approximately Normal. Normal Condition for Sample Means If the population distribution is Normal, then so is the sampling distribution of This is true no matter what the sample size n is. If the population distribution is not Normal, the central limit theorem tells us that the sampling distribution of will be approximately Normal in most cases if n ≥ 30.

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Young Women’s Heights The height of young women follows a Normal distribution with mean μ = 64.5 inches and standard deviation σ = 2.5 inches. Find the probability that a randomly selected young woman is taller than 66.5 inches. Show your work. (b) Find the probability that the mean height of an SRS of 10 young women exceeds 66.5 inches. Show your work. P(X > 66.5) = P(z > 0.80) = P(X > 66.5) = P(z > 2.53) =

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**Buy Me Some Peanuts and Sample Means**

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. (b) Find the probability of each event described above. (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. P(X<16) = P( <16) =

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**Check Your Understanding**

The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Suppose we choose an SRS of 6 pregnant women. Let X = the mean pregnancy length for the sample. 2. What is the mean of the sampling distribution of ? Explain. 1. Find the probability that a randomly chosen pregnant woman has a pregnancy that lasts for more than 270 days. Show your work. 266 Days 0.4013 3. Compute the standard deviation of the sampling distribution of Check that the 10% condition is met. 6.532 days 4. Find the probability that the mean pregnancy length for the women in the sample exceeds 270 days. Show your work. 0.2709

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**Servicing Air Conditioners**

Your company has a contract to perform preventive maintenance on thousands of air-conditioning units in a large city. Based on service records from the past year, the time (in hours) that a technician requires to complete the work follows the distribution whose density curve is shown. This distribution is strongly right-skewed, and the most likely outcomes are close to 0. The mean time is μ = 1 hour and the standard deviation is σ = 1 hour. In the coming week, your company will service an SRS of 70 air-conditioning units in the city. You plan to budget an average of 1.1 hours per unit for a technician to complete the work. Will this be enough?

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Answer: 10% condition is met Approximately Normal shape since the Normal condition is met: n = 70 > 30 P(X > 1.1) = P(z > 0.83) = 1 − =

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

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Exercises on page 441, #43-46, & Page 454 #49, 51, 53, 55

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