Examples for Week 5Homework Problems7,9,13,15,17 and 20

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Example for Week 5Homework Problem 7

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Homework Example

• Assume the random variable x is normally distributed with mean mu=73 and standard deviation sigma = 6. Find the indicated probability– P(x< 66)

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• Minitab– Graph >> Probability Distribution Plot

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After choosing Normal Distribution and inputting mean and standard deviation, click Shaded Area Tab

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Left tailed because you are finding the probability of P(x<66)…Remember “<“ points to the left

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Answer

Example for Week 5Homework Problem 9

B Heard

Homework Example

• A survey was conducted to measure the height of men. In the survey, respondents were grouped by age. In the 20-29 group, the heights were normally distributed with a mean of 68.7 inches and a standard deviation of 3.1 inches. A study participant is randomly selected. Find– Probability that his height is less than 68 inches– Probability his height is between 68 and 71 inches– Probability his height is more than 70 inches

Homework Example

• Minitab– Graph >> Probability Distribution Plot

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After choosing Normal Distribution and inputting mean and standard deviation, click Shaded Area Tab. This will stay the same for all three parts of this problem.

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Left tailed because you are finding the probability of P(x<68)…Remember “<“ points to the left

Probability that his height is less than 68 inches

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Answer

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• For the next part - Same process, we will only change the choices on the Shaded Area Tab

Homework ExampleProbability his height is between 68 and 71 inches

Choose “Middle” since you are finding the probability between two values.

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Answer

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• For the next part - Same process, we will only change the choices on the Shaded Area Tab

Homework ExampleProbability his height is more than 70 inches

Right Tailed because you are finding P(x>70), “>” points to the right.

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Answer

Example for Week 5Homework Problem 13

B Heard

Homework Example

• Find the z-score that has the 79.5% of the distribution’s area to its right.

• THIS IS EASY – Remember the z-score/Standard Normal Tables are based on a mean of 0 and a Standard Deviation of 1

• USE Minitab

Homework Example

• Minitab– Graph >> Probability Distribution Plot

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Dealing with z-scores? Use a mean of 0 and standard deviation of 1 (and set Distribution to Normal)

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Click the radial button next to PROBABILITY, because we are inputting a probability, Right Tail because we are looking for the z-score with that has 79.5% TO ITS RIGHT, Change 79.5% to decimal form (0.795)

Find the z-score that has the 79.5% of the distribution’s area to its right.

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Note Answer is on x axis since we input the probability.Answer is -0.82 to two decimal places (Don’t forget the negative sign)

Example for Week 5Homework Problem 15

B Heard

Homework Example

• A survey was conducted to measure the height of men. In the survey, respondents were grouped by age. In the 20-29 group, the heights were normally distributed with a mean of 68.7 inches and a standard deviation of 3.1 inches. A study participant is randomly selected. Find– The height that represents the 80th percentile– The height that represents the first quartile

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• Minitab– Graph >> Probability Distribution Plot

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Just input your mean, standard deviation and choose normal distribution.

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Click the radial button next to PROBABILITY, because we are inputting a probability, Left Tail because 80th percentile means 80% of the area is to the left, 80% is 0.80 in decimal form…. Bazinga!

You are looking for the height that represents the “80th Percentile”

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Note Answer is on x axis since we input the probability.Answer is 71.31 to two decimal places (80% would be below this height)

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Click the radial button next to PROBABILITY, because we are inputting a probability, Left Tail because first quartile 25% of the area is to the left, 25% is 0.25 in decimal form…. Bazinga!

You are looking for the “first quartile” which is really just the “25th Percentile”

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Note Answer is on x axis since we input the probability.Answer is 66.61 to two decimal places (25% would be below this height)

Example for Week 5Homework Problem 17

B Heard

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• A population has a mean mu=72 and a standard deviation sigma =16. Find the mean and standard deviation of a sampling distribution of sample means with sample size n=64.

Homework Example

• A population has a mean mu=72 and a standard deviation sigma =16. Find the mean and standard deviation of a sampling distribution of sample means with sample size n=64.

This “SCREAMS” – “Youhave to recalculate

standard deviation!”

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72

Why? Remember the screaming?2To get the standard deviation of sampling distribution you must divide the standard deviation of the population by the square root of the sample size16/square root (64) = 16/8 = 2

Same as Population Mean

Just remember the screaming – it’s easy!

Example for Week 5Homework Problem 20

B Heard

Homework Example

• A tire company claims the life span of its tires is 60,000 miles. You work for a consumer research company and are assigned to test the tires. Assume the life span of the tires is normally distributed. You select 100 tires and test them. The mean life of the tires is 59,590 miles. Assume the population standard deviation sigma = 900. If you assume the company’s claim is correct, find– Probability that the mean of the sample is 59,590 miles

or less

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• This would should SCREAM at you because you were given a sample size of “100”– Thus you will have to change your standard

deviation• It would be the population standard deviation divided

by the square root of the sample size or 900/square root (100) = 900/10 = 90• The rest is similar to some of the other problems, but I

will show you

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• Minitab– Graph >> Probability Distribution Plot

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After choosing Normal Distribution and inputting mean and recalculated standard deviation, click Shaded Area Tab.

Company’s Claim

Recalculated Standard Deviation

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Left tailed because you are finding the probability of P(x<59590)…Remember “<“ points to the left

Probability that his height is less than 68 inches

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A very small number…2.6124E-6 means2.6124 x 10-6 or0.0000026124(Yours won’t be this small probably)

Answer

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• For My Example, I could say…– The claim is inaccurate because the sample mean

would be considered unusual since it does not lie within the range of a usual event, namely within 2 standard deviations of the mean of the sample means. (It’s a lot farther away than 2 standard deviations)

– However, assuming the company’s claim is true, this would NOT be unusual because 59,590 does lie with the range of a usual event , namely 2 standard deviations of the mean for an individual tire. (60,000 +/- 2(900)) or between 58,200 and 61,800.