Week 7 LectureStatistics for Decision

MakingB. Heard

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Standard Normal Distribution◦ The “standard” normal distribution is a normal

distribution with mean zero and where the standard deviation (and variance) equals one.

◦ The Total Area under the curve is one (1) or 100% (This is true for all normal distributions regardless of the mean and standard deviation).

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Using Minitab for Normal Distribution calculations.

Use Calc >> Probability Distributions >> Normal

Examples Follow

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Example

The average fish in Happy Lake weighs 2 pounds with a standard deviation of 0.5 pounds. If Bob catches a fish that weighs 3.2 pounds. What could you say about the catch?

Week 7 Lecture

Week 7 Lecture

Since this is the Cumulative Distribution Function, it “fills” from left to right. Therefore, you could say his catch was in the “Top 1 %”

Example

The average fish in Happy Lake weighs 2 pounds with a standard deviation of 0.5 pounds. If Bob catches a fish that weighs 1.35 pounds. What could you say about the catch?

Week 7 Lecture

Week 7 Lecture

Since this is the Cumulative Distribution Function, it “fills” from left to right. Therefore, you could say his catch was in the “Bottom 10 %”

Other types of questions If you have a normal distribution with a mu

= 100 and sigma = 15, what number corresponds to a z = -2

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-2 = (x – 100)/15Multiply both sides by 15 to get-30 = x – 100Add 100 to each side to get70 = x

So “70” is my answer, I just did a little Algebra.

Another type of question Say we take 120 samples of size 81 each

from a distribution we know is normal. Calculate the standard deviation of the sample means if we know the population variance is 25.

(Answer next chart)

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Answer The Central Limit Theorem tells us the

variance is the Population variance divided by the Sample Size. We can just take the square root to get the standard deviation.

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Variance = 25/81 or 0.309Standard Deviation = Square Root(25/81) = 5/9 = 0.556

Finding z scores Example

◦ The area to the left of the “z” is 0.6262. What z score corresponds to this area.

◦ Use Calc >> Probability Distributions >> Normal◦ (Set Mean = 0 and Standard Deviation to 1 and

use “INVERSE Cumulative Probability”

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Week 7 Lecture

Answer is 0.322 rounded to three decimals. Remember the distribution fills from left to right.

Another type of question In a normal distribution with mu = 40 and

sigma = 10 find P(32 < x < 44)◦ Easy, but this takes a couple of steps.◦ Using Calc >> Probability Distributions >>

Normal find the probabililties that x < 32 and x < 44 using the Cumulative Probability option.

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Continued

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Get results for Both 32 and then 44.

Answer

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Subtract 0.655422 – 0.211855To get0.443567Or 0.444 rounded tothree decimals

P(32 < x < 44) = 0.444 based onthe given mean and std deviation.

Confidence Intervals and Examples◦ Charts follow

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Interpreting Confidence Intervals◦ If you have a 90% confidence interval of (15.5,

23.7) for a population mean, it simply means “There is a 90% chance that the population mean is contained in the interval (15.5, 23.7) It’s really that simple.

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Finding Confidence Intervals◦ A luxury car company wants to estimate the true

mean cost of its competitor’s automobiles. It randomly samples 180 of its competitors sticker prices. The mean cost is $65,000 with a standard deviation of $3200. Find a 95% confidence interval for the true mean cost of the competitor’s automobiles. Write a statement about the interval.

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It randomly samples 180 of its competitors sticker prices. The mean cost is $65,000 with a standard deviation of $3200. Find a 95% confidence interval…

Use Stat >> Basic Statistics >> 1 sample Z◦ Make sure to click Options and set to 95%

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Week 7 Lecture

Click your OK buttons…

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Confidence Interval is (64533, 65467), which means we can be 95% confident the true mean cost of the competitor’s vehicles are between those two values.

Find Confidence Intervals of Proportions Example

◦ An student wants to estimate what proportion of the student body eats on campus. The student randomly samples 200 students and finds 120 eat on campus. Using a 95% confidence interval, estimate the true proportion of students who eat on campus. Write a statement about the confidence level and interval.

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Example Solution◦ p hat = 120/200 = 0.60◦ q hat = 1- 0.60 = 0.40◦ n p hat = 200 * 0.60 = 120◦ n q hat = 200 * 0.40 = 80◦ Using E = Zc* Square Root ((p hat * q hat)/n) = 1.96 * Square Root ((0.60*0.40)/200)=0.0679Now we subtract this from the mean for the left side of the

interval and add it to the mean for the right side. (0.60 – 0.0679, 0.60 + 0.0679) = (0.5321, 0.6679)

So with 95% confidence, we can say the population proportion of students who eat lunch on campus is (0.5321, 0.6679) or between 53.21% and 66.79%.

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Link to charts will be posted at

◦ www.facebook.com/statcave

PLEASE NOTE THAT I WILL BE BACK HERE NEXT SUNDAY NIGHT FOR A BONUS LECTURE TO HELP YOU PREPARE FOR THE FINAL EXAM.

ENTER IN THE WEEK 7 iConnect AREA JUST LIKE YOU DID TONIGHT.

Week 7 Lecture