1 / 35

Working With Normal Models

Working With Normal Models. Driving. It takes you 20 minutes, on average, to drive to school with a standard deviation of 2 minutes Suppose a Normal Model is appropriate for the distribution of driving times. A) How often will you arrive at school in less than 22 minutes? Answer:

glyn
Download Presentation

Working With Normal Models

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Working With Normal Models

  2. Driving • It takes you 20 minutes, on average, to drive to school with a standard deviation of 2 minutes • Suppose a Normal Model is appropriate for the distribution of driving times • A) How often will you arrive at school in less than 22 minutes? • Answer: 68% of the time we’ll be within 1 SD, or two minutes, of the average 20 minutes. So 32% of the time we’ll arrive in less than 18 minutes or in more than 22 minutes. Half of those times (16%) will be greater than 22 minutes, so 84% will be less than 22 minutes

  3. Driving • It takes you 20 minutes, on average, to drive to school with a standard deviation of 2 minutes • Suppose a Normal Model is appropriate for the distribution of driving times • B) How often will it take you more than 24 minutes? • Answer: 24 minutes is 2 SD above the mean. By the 95% rule, we know 2.5% of the times will be more than 24 minutes

  4. Driving • It takes you 20 minutes, on average, to drive to school with a standard deviation of 2 minutes • Suppose a Normal Model is appropriate for the distribution of driving times • C) Do you think the distribution of your driving times is unimodal and symmetric? • Answer: “Good” traffic will speed up your time by a bit but traffic incidents may occasionally increase the time it takes so times may be skewed to the right and there may be outliers.

  5. Driving • It takes you 20 minutes, on average, to drive to school with a standard deviation of 2 minutes • Suppose a Normal Model is appropriate for the distribution of driving times • D) What does the shape of the distribution then say about the accuracy of your predictions? • Answer: If this is the case the Normal Model is not appropriate and the percentages we predict would not be accurate.

  6. Finding Normal Percentiles

  7. SAT Scores Example • Each 800 point section of the SAT exam is constructed to have a mean of 500 and a Standard Deviation of 100 • The distribution of scores is unimodal and symmetric • Suppose you earned a 600 on one part of the SAT, where do you fall among all students who took the test?

  8. SAT Scores Example • Does the z-Score tell you anything? • We need to employ our Normal Model!

  9. SAT Scores Example • 68% of students had scores that fell no more than 1 SD from the mean. • 100% - 68% = 32% had scores more than 1 standard deviation from the mean • Only half of those were on the high side, so 16% of students were better than my score of 600. • My score is higher than about 84% of students taking the exam.

  10. Finding Normal Percentiles i.e. what if your score was 680?

  11. We Have Two Options: • If the value doesn’t fall at exactly 1, 2, or 3 SDs from the mean, we have two options: • Technology • Tables Either way, start by standardizing into z-scores

  12. Table • To Find In Table: • Look down left column for the first two digits, 1.8 • Then across the row for the third digit, 0 The table gives .9641, this means 96.41% of the z-scores are less than 1.80. Only 3.6% of people, then, scored better than 680 on the SATs

  13. Calculator! • Look under 2nd DISTR • There are three “norm” functions:normalpdf(normalcdf(invNorm( • normalpdf(calculatesy-values for graphing a Normal Curve • We won’t use this one much but lets try it now, Y1=normalpdf(X) in a graphing WINDOW with Xmin=-4, Xmax=4, Ymin=-0.1, and Ymax=0.5

  14. Calculator! • Normalcdf( finds the proportion of area under the curve between two z-score cut points, by specifying normalcdf(zLeft, zRight) • You will use this function often!

  15. Calculator! Example • Example 1 • Let’s find the shaded area: • Under 2nd DISTR: • Select normalcdf( • Hit ENTER • Specify the cut points Normalcdf(-.5,1) and hit ENTER You get 0.533, and that is the area between those two points, 53.3% Approximately 53% of a Normal Model falls between half a standard deviation and below 1 standard deviation above the mean

  16. Example 2 • Previous SAT example: • We determined the fraction of scores above your score of 680

  17. SAT Example • First we need to find z-scores for the cut points: • 680 is 1 SD above the mean, your z-score is 1.8This is the left cut-point • The standard Normal extend rightward forever, but remember how little of the data lies beyond +/- 3 SD • Take the upper cut point of say, 99. Use +/-99 as your general rule of thumb • Use the command: Normalcdf(1.8,99) Answer: .0359302655 So 3.6% of SAT scores are higher than 680.

  18. From Percentiles to z-Scores

  19. Typical Question • What z-score cuts off the top 10% in a Normal Model? • Before – find areas (percentages of total) from the z-score • Now – Find z-score from percentages

  20. By Table • From the table, we need an area of .900 • This exact area is not there, but .8997 is pretty close • This shows up in table with 1.2 in left margin and .08 in top margin • Z-score for 90th percentile is 1.28

  21. By Calculator • The function we will use is under 2nd DISTR -> invNorm( • For the top 10%, we take invNorm(.90) and get: • 1.281551567 • Only 10% of the area in a Normal Model is more than about 1.28 SDs above the mean

  22. Pesky Word Problems! • (pg121) • A cereal manufacturer has a machine that fills cereal boxes • Boxes are labeled 16oz • The machine is never going to be perfect so there will be minor variations • If the machine is set to 16oz and the Normal model applies, about half the boxes will be underweight, leading to angry customers

  23. Pesky Cereal Problem! • THINK Question: What portion of cereal boxes will be under 16 oz? Variable: Let y = weight of each cereal box Use a N(16.3,0.2) model

  24. Pesky Word Problems! • To avoid possible lawsuits from angry customers, the company sets the mean to be a little higher than 16oz • The company believes the packaging machine fills in an amount of cereal that fits a Normal model with a SD = .2oz • The company decides to put an average of 16.3oz of cereal in each box • Question 1: What fraction of cereal boxes will be underweight?

  25. Cereal!!! • Convert cut-off value into a z-score: • Find the area with your calculator: • Area(y<16)=Area(z<-1.5)=.0668

  26. Cereal… • Question 1: What percent of cereal boxes will be underweight? • Answer 1: Approximately 6.7% of the boxes will contain less than 16oz

  27. Cereal Question 2 • The company’s lawyers say that 6.7% is too high. They insist that no more than 4% of cereal boxes be underweight. • What new mean setting does the company need to insure this happens? • Let y=weight of cereal box • I don’t know the mean amount of cereal. The standard deviation is still 0.2 oz, so the model is

  28. Cereal Q2 • Solution: • Find a z-score that cuts off the lowest 4% • Use this information to find . • The z-score with .04 area to the left is z= -1.75 • What function finds this?

  29. Cereal Q2 • For 16 to be -1.75 SD below the mean, the mean must be: • 16+1.75(.02)=16.35oz • The company must set the machine to average 16.35oz of cereal per box to have less than 4% be underweight

  30. Cereal……….Question 3! • The President of the cereal company isn’t happy. He thinks they should give away less free cereal, not more. • His goal is to set the machine no higher than 16.2oz and still have only 4% underweight boxes • The only way to accomplish this is to reduce the standard deviation. What SD must the company achieve and what does this mean about the machine?

  31. Cereal 3 • Question: What SD will allow the mean to be 16.2oz and still have only 4% of boxes underweight? • Let y = weight cereal in box • I know the mean, but not the SD, so use

  32. Cereal 3 • I know the z-score with 4% below it is -1.75 from previous problem • Sole for

  33. Cereal 3 • The company must get the machine to box cereal with a standard deviation of only 0.114 ounces. This means the machine must be more consistent (by nearly a factor of 2) at filling the boxes.

  34. Homework Pg 130, # 17, 19, 21, 37, 41

More Related