Working With Normal Models

Working With Normal Models

DrivingIt 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 timesA) How often will you arrive at school in less than 22 minutes?

Answer:68% of the time well be within 1 SD, or two minutes, of the average 20 minutes.So 32% of the time well 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 minutesDrivingIt 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 timesB) 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 minutesDrivingIt 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 timesC) 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.DrivingIt 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 timesD) 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.Finding Normal PercentilesSAT Scores ExampleEach 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 symmetricSuppose you earned a 600 on one part of the SAT, where do you fall among all students who took the test?SAT Scores Example

SAT Scores Example68% 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.

Finding Normal Percentilesi.e. what if your score was 680?We Have Two Options:If the value doesnt fall at exactly 1, 2, or 3 SDs from the mean, we have two options:TechnologyTables

Either way, start by standardizing into z-scores

TableThe 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 SATsCalculator!Look under 2nd DISTRThere are three norm functions:normalpdf(normalcdf(invNorm(normalpdf( calculatesy-values for graphing a Normal CurveWe wont 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

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!

Calculator! ExampleExample 1Lets find the shaded area:

Under 2nd DISTR:Select normalcdf(Hit ENTERSpecify the cut pointsNormalcdf(-.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

Example 2Previous SAT example:

We determined the fraction of scores above your score of 680

SAT ExampleFirst 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-pointThe standard Normal extend rightward forever, but remember how little of the data lies beyond +/- 3 SDTake the upper cut point of say, 99. Use +/-99 as your general rule of thumbUse the command:

Normalcdf(1.8,99)

Answer: .0359302655

So 3.6% of SAT scores are higher than 680.From Percentiles to z-ScoresTypical QuestionWhat z-score cuts off the top 10% in a Normal Model?

Before find areas (percentages of total) from the z-scoreNow Find z-score from percentages

By TableFrom 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

By CalculatorThe 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

Pesky Word Problems!(pg 121)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

Pesky Cereal Problem!THINKQuestion: 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

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 = .2ozThe company decides to put an average of 16.3oz of cereal in each box

Question 1: What fraction of cereal boxes will be underweight?Cereal!!!

CerealQuestion 1: What percent of cereal boxes will be underweight?

Answer 1: Approximately 6.7% of the boxes will contain less than 16oz

Cereal Question 2The companys 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?Cereal Q2

Cereal Q2For 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

Cereal.Question 3!The President of the cereal company isnt 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?

Cereal 3

Cereal 3Cereal 3The 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.HomeworkPg 130, # 17, 19, 21, 37, 41