Sapling 7.3

The number of hours a light bulb burns before failing varies from bulb to bulb. The population distribution of burnout times is strongly skewed to the right. The central limit theorem says that

the average burnout time of a large number of bulbs has a sampling distribution that is close to Normal.

A bottling company uses a filling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 milliliters (ml). In fact, the contents vary according to a Normal distribution with mean 𝜇μ = 298 ml and standard deviation 𝜎σ = 3 ml. What is the probability that the mean contents of six randomly selected bottles is less than 295 ml?

0.0072

An insurance company claims that in the entire population of homeowners, the mean annual loss from fire is 𝜇μ = $250 and the standard deviation of the loss is 𝜎σ = $5000. The distribution of losses is strongly right-skewed: many policies have $0 loss, but a few have large losses. The company hopes to sell 1000 of these policies for $300 each. If the company wants to be 90% certain that the mean loss from fire in an SRS of 1000 homeowners is less than the amount it charges for the policy, how much should the company charge?

$452.63

The level of cholesterol in the blood for all men aged 20 to 34 follows a Normal distribution with mean 𝜇𝑀=188μM=188 milligrams per deciliter (mg/dl) and standard deviation 𝜎𝑀=41σM=41 mg/dl. For 14-year-old boys, blood cholesterol levels follow a Normal distribution with mean 𝜇𝐵=170μB=170 mg/dl and standard deviation 𝜎𝐵=30σB=30 mg/dl. Suppose we select independent SRSs of 25 men aged 20 to 34 and 36 boys aged 14 and calculate the sample mean cholesterol levels 𝑥¯𝑀x¯M and 𝑥¯𝐵x¯B. -Is the sample size of 14-year old boys less than 10% of all 14-year-old boys? -Is the sample size of men aged 20 to 34 less than 10% of all men aged 20 to 34? -Is the 10% condition met? Calculate and interpret the standard deviation of the sampling distribution of 𝑥¯𝑀−𝑥¯𝐵x¯M−x¯B?

- yes - yes -yes 9.60 mg/dl. The difference (20- to 34-year old males - 14-year old boys) in the sample mean blood cholesterol levels typically varies by about 9.60 mg/dl from the true difference in means of 18 mg/dl.

The number of flaws per square yard in a type of carpet material varies with mean 1.6 flaws per square yard and standard deviation 1.2 flaws per square yard. Calculate the probability of the second event in randomly selecting 50 square yards of material and finding an average of 2 or more flaws

0.0092

A car company claims that the lifetime of its batteries varies from car to car according to a Normal distribution with mean 𝜇μ= 48 months and standard deviation 𝜎σ = 8.2 months. A consumer organization installs this type of battery in an SRS of 8 cars and calculates 𝑥¯x¯ = 42.2 months. Find the probability that the sample mean lifetime is 42.2 months or less if the company's claim is true.

0.0227

The level of cholesterol in the blood for all men aged 20 to 34 follows a Normal distribution with mean 𝜇𝑀=188μM=188 milligrams per deciliter (mg/dl) and standard deviation 𝜎𝑀=41σM=41 mg/dl. For 14-year-old boys, blood cholesterol levels follow a Normal distribution with mean 𝜇𝐵=170μB=170 mg/dl and standard deviation 𝜎𝐵=30σB=30 mg/dl. Suppose we select independent SRSs of 25 men aged 20 to 34 and 36 boys aged 14 and calculate the sample mean cholesterol levels 𝑥¯𝑀x¯M and 𝑥¯𝐵x¯B. What is the probability of getting a difference in sample means 𝑥¯𝑀−𝑥¯𝐵x¯M−x¯B that's less than 0 mg/dl?Round to 4 decimal places.

0.0304

A bottling company uses a filling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 milliliters (ml). In fact, the contents vary according to a Normal distribution with mean 𝜇μ = 298 ml and standard deviation 𝜎σ = 3 ml. What is the probability that the contents of a randomly selected bottle is less than 295 ml?

0.1587

An insurance company claims that in the entire population of homeowners, the mean annual loss from fire is 𝜇μ = $250 and the standard deviation of the loss is 𝜎σ = $5000. The distribution of losses is strongly right-skewed: many policies have $0 loss, but a few have large losses. The company hopes to sell 1000 of these policies for $300 each. Assuming that the company's claim is true, what is the probability that the mean loss from fire is greater than $300 for an SRS of 1000 homeowners?

0.3759

A machine is designed to fill 16-ounce bottles of shampoo. When the machine is working properly, the amount poured into the bottles follows a Normal distribution with mean 16.05 ounces and standard deviation 0.1 ounce. Assume that the machine is working properly. If 4 bottles are randomly selected and the number of ounces in each bottle is measured, then there is about a 95% probability that the sample mean will fall in which of the following intervals?

15.95 to 16.15 ounces

The level of cholesterol in the blood for all men aged 20 to 34 follows a Normal distribution with mean 𝜇𝑀=188μM=188 milligrams per deciliter (mg/dl) and standard deviation 𝜎𝑀=41σM=41 mg/dl. For 14-year-old boys, blood cholesterol levels follow a Normal distribution with mean 𝜇𝐵=170μB=170 mg/dl and standard deviation 𝜎𝐵=30σB=30 mg/dl. Suppose we select independent SRSs of 25 men aged 20 to 34 and 36 boys aged 14 and calculate the sample mean cholesterol levels 𝑥¯𝑀x¯M and 𝑥¯𝐵x¯B. What is the mean of the sampling distribution of 𝑥¯𝑀−𝑥¯𝐵x¯M−x¯B?

18 mg/dl

David's iPod has about 10,000 songs. The distribution of the play times for these songs is heavily skewed to the right with a mean of 225 seconds and a standard deviation of 60 seconds. Suppose we choose an SRS of 10 songs from this population and calculate the mean play time x of these songs. How many songs would you need to sample if you wanted the standard deviation of the sampling distribution of 𝑥¯x¯ to be 10 seconds? Justify your answer.

36

David's iPod has about 10,000 songs. The distribution of the play times for these songs is heavily skewed to the right with a mean of 225 seconds and a standard deviation of 60 seconds. Describe the shape of the sampling distribution of 𝑥¯x¯ for SRSs of size 𝑛=5n=5 from the population of songs on David's iPod. Justify your answer.

Because n = 5 < 30, the sampling distribution of 𝑥¯x¯ will also be skewed to the right, but not quite as strongly as the population.

Administrators at a small school with 200 students want to estimate the average amount of time students spend looking at a screen (phone, computer, television, and so on) per day. The administrators select a random sample of 50 students from the school to ask. What effect does violating the 10% condition have on the standard deviation of the sampling distribution of ¯xx¯?

Because the administrators surveyed more than 10% of the finite population (of 200 students), the actual standard deviation of the sampling distribution of ¯xx¯ will be less than the value provided by the formula σ¯x=σ√nσx¯=σn.

A car company claims that the lifetime of its batteries varies from car to car according to a Normal distribution with mean 𝜇μ= 48 months and standard deviation 𝜎σ = 8.2 months. A consumer organization installs this type of battery in an SRS of 8 cars and calculates 𝑥¯x¯ = 42.2 months. The probability 𝑃(𝑥¯≤42.2)P(x¯≤42.2) was found to be 0.0227. Based on this probability, is there convincing evidence that the company is overstating the average lifetime of its batteries?

Because this probability is very small, there is convincing evidence that the company is overstating the average lifetime of its batteries. It is not plausible to get a sample mean this small by chance alone.

David's iPod has about 10,000 songs. The distribution of the play times for these songs is heavily skewed to the right with a mean of 225 seconds and a standard deviation of 60 seconds. Describe the shape of the sampling distribution of 𝑥¯x¯ for SRSs of size 𝑛n = 100 from the population of songs on David's iPod. Justify your answer.

Because 𝑛=100≥30n=100≥30, the sampling distribution of 𝑥¯x¯ is approximately Normal by the central limit theorem.

The number of flaws per square yard in a type of carpet material varies with mean 1.6 flaws per square yard and standard deviation 1.2 flaws per square yard. Without doing any calculations, explain which event is more likely: • randomly selecting a 1 square yard of material and finding 2 or more flaws • randomly selecting 50 square yards of material and finding an average of 2 or more flaws

It is more likely to randomly select 1 square yard of material and find 2 or more flaws. There is more variability from the mean of 1.6 in the number of flaws found in individual square yards of material than in the average number of flaws found in a sample of 50 square yards of material.

The distribution of scores on the mathematics part of the SAT exam in a recent year was approximately Normal with mean 515 and standard deviation 114. Imagine choosing many SRSs of 100 students who took the exam and averaging their SAT Math scores. Which of the following are the mean and standard deviation of the sampling distribution of 𝑥¯x¯?

Mean = 515, SD = 114/√100

Asked what the central limit theorem says, a student replies, "As you take larger and larger samples from a population, the histogram of the sample values looks more and more Normal." Is the student right? Explain your answer.

No. The histogram of the sample values will look like the population distribution, whatever it might happen to be.

Administrators at a small school with 200 students want to estimate the average amount of time students spend looking at a screen (phone, computer, television, and so on) per day. The administrators select a random sample of 50 students from the school to ask.

The administrators sampled 25% of the students from the school. (Do not round) The 10% condition is not met because 50 is not less than 10% of all students at the small school.

The level of cholesterol in the blood for all men aged 20 to 34 follows a Normal distribution with mean 𝜇𝑀=188μM=188 milligrams per deciliter (mg/dl) and standard deviation 𝜎𝑀=41σM=41 mg/dl. For 14-year-old boys, blood cholesterol levels follow a Normal distribution with mean 𝜇𝐵=170μB=170 mg/dl and standard deviation 𝜎𝐵=30σB=30 mg/dl. Suppose we select independent SRSs of 25 men aged 20 to 34 and 36 boys aged 14 and calculate the sample mean cholesterol levels 𝑥¯𝑀x¯M and 𝑥¯𝐵x¯B. What is the shape of the sampling distribution of 𝑥¯𝑀−𝑥¯𝐵x¯M−x¯B? Why?

The shape of the distribution of 𝑥¯𝑀−𝑥¯𝐵x¯M−x¯B is Normal because both population distributions are Normal.

Why is it important to check the 10% condition before calculating probabilities involving 𝑥¯x¯?

To ensure that the observations in the sample are close to independent

The level of cholesterol in the blood for all men aged 20 to 34 follows a Normal distribution with mean 𝜇𝑀=188μM=188 milligrams per deciliter (mg/dl) and standard deviation 𝜎𝑀=41σM=41 mg/dl. For 14-year-old boys, blood cholesterol levels follow a Normal distribution with mean 𝜇𝐵=170μB=170 mg/dl and standard deviation 𝜎𝐵=30σB=30 mg/dl. Suppose we select independent SRSs of 25 men aged 20 to 34 and 36 boys aged 14 and calculate the sample mean cholesterol levels 𝑥¯𝑀x¯M and 𝑥¯𝐵x¯B. The probability of getting a difference in sample means 𝑥¯𝑀x¯M and 𝑥¯𝐵x¯B that's less than 0 mg/dl is 0.0301. Should we be surprised if the sample mean cholesterol level for the 14-year-old boys exceeds the sample mean cholesterol level for the men? Explain your answer.

Yes. The likelihood that the sample mean cholesterol level of the boys is greater than the sample mean cholesterol level of the men is only 3%.

The number of flaws per square yard in a type of carpet material varies with mean 1.6 flaws per square yard and standard deviation 1.2 flaws per square yard. Without doing any calculations, explain why you cannot use a Normal distribution to calculate the probability of randomly selecting a 1 square yard of material and finding 2 or more flaws.

You cannot use a Normal distribution to calculate the probability of randomly selecting a 1 square yard of material and finding 2 or more flaws because the population distribution is not Normal and the sample size is not at least 30.

David's iPod has about 10,000 songs. The distribution of the play times for these songs is heavily skewed to the right with a mean of 225 seconds and a standard deviation of 60 seconds. Suppose we choose an SRS of 10 songs from this population and calculate the mean play time x of these songs. Identify the mean of the sampling distribution of 𝑥¯x¯.

𝜇𝑥¯=𝜇μx¯=μ = 225

David's iPod has about 10,000 songs. The distribution of the play times for these songs is heavily skewed to the right with a mean of 225 seconds and a standard deviation of 60 seconds. Suppose we choose an SRS of 10 songs from this population and calculate the mean play time 𝑥¯x¯ of these songs. Calculate and interpret the standard deviation of the sampling distribution of 𝑥¯x¯.

𝜎𝑥¯=𝜎𝑛√=6010√=σx¯=σn=6010= 18.97 secondsIn SRSs of size 10, the sample mean play time will typically vary by about 18.97 seconds from the true mean of 225 seconds.