3by9on w week_5

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

B Heard

Example for Week 5Homework Problem 7

B Heard

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)

Homework Example

• Minitab– Graph >> Probability Distribution Plot

Homework Example

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

Homework Example

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

Homework Example

Answer


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


Homework Example

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.

Homework Example


Probability that his height is less than 68 inches

Homework Example

Answer

Homework Example

• 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.

Homework Example

Answer

Homework Example

• 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.

Homework Example

Answer


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


Homework Example

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

Homework Example

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.

Homework Example

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)


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

Homework Example


Homework Example

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

Homework Example

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”

Homework Example

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)

Homework Example

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”

Homework Example

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)


B Heard

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.

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!”

Homework Example

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!


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

Homework Example

• 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

Homework Example


Homework Example

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

Company’s Claim

Recalculated Standard Deviation

Homework Example


Probability that his height is less than 68 inches

Homework Example

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

Answer

Homework Example

• 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.

3by9on w week_5

Design

Transcript of 3by9on w week_5