The Normal Models/Distributions

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Normal Distribution ParametersThe first parameter specified here is shape: symmetric, unimodal, bell shaped. (A formula defines the exact shape.)

For a Normal variable the mean and standard deviation are needed to proceed.

All else follows from knowing these three things.

Z-scores (Standardized Values)

StDevMeanXZ

Z measures“how many standard deviations from the mean”

whether above (+) or below (-) the mean

This is the same formula, solved for X.

You are required to know these formulas.

StDevZMeanX

ExampleThe gestation period (length of pregnancy) for women is Normally distributed with mean = 270 days and standard deviation = 11 days.

Draw a picture!

ExampleUse the mean to get started.

ExampleOn the templates, consecutive tick marks are one standard deviation apart.

303292281270259248237

Length of Pregnancy (Days)

Example

A reference to getting started with Normal distributions.

About 68% of the data fall within 1 standard deviation of the mean:

68 – 95 – 99.7 Rulealso known as the

Empirical Rule

Example

0.6827

Empirical (68 – 95 – 99.7) RuleA reference to getting started with Normal distributions.

About 95% (all but roughly 1 out of 20) of the data fall within 2 standard deviations of the mean: 2

Example

0.9544

Empirical (68 – 95 – 99.7) RuleA reference to getting started with Normal distributions.

About 99.7% (all but roughly 1 in 370) of the data fall within 3 standard deviations of the mean: 3

Example

0.9973

Data more than 3 from 0.0027 or about 1 in 370 data values (on average) fall outside 3.

Technically the minimum and maximum are - and + (infinite). Values far to the right and left have very very very very small likelihood.

Beyond + 3

0.00135 =1/741 values more than 3 above

0.0000317 =1/31574 values more than 4 above

If the curve were drawn 1 mm high at 6 from , then it would be drawn 65.66 km high at .

ExampleThe Z scale “goes with” the data scale.

Z = # of standard deviations from the mean.

ExampleDetermine the probability a pregnancy lasts at least 274 days (that’s 9 months).

Finding ProbabilitiesTo solve a problem asking for a probability with for a Normal distribution:

Step 1. Draw a picture identifying the relevant area.

ExampleDetermine the probability a pregnancy lasts at least 274 days.

Area = Probability

Step 2. Convert the data value to the Z scale.

Data = 274 = 270 = 11 days

274 is 0.36 standard deviations above the mean

36.011

270274

Step 3. Use the Table (A-2 in the book) or computer or calculator to determine the percentile for the value.

A Z-score of 0.36 matches with a probability of 0.6406. That is: Z = 0.36 is the 64.06 percentile.

0.6406

Step 4. Problems asking for the probability of a result above or between will require further operations, because of the way percentiles are oriented.

Example

0.6406 1 – 0.6406

= 0.3594

coincidence

Example - Solution

0.3594 is the probability that a pregnancy lasts at least 274 days.

That is:35.94% of all pregnancies last at least 274 days. (Probability calculations always apply to populations.)

274 days is the 64 percentile of gestation times.

ExampleThe gestation period (length of pregnancy) for women is Normally distributed with mean = 270 days and standard deviation = 11 days.

ExampleA women gave birth to a child on October 2, 2007. She identifies a man as the father. The case is in court.

The man testifies he is not the father, and documents that he was out of the country from December 19, 2006 to January 30, 2007.

December 19 is 287 days before October 2.

January 30 is 245 days before October 2.

What is the probability a pregnancy lasts between 245 and 287 days?

(If it’s high, then we have evidence – in the data – supporting the man’s claim that he is not the father.)

ExampleWhat’s the probability the pregnancy lasted between 245 and 287 days?

Area = Probability

Convert both data values to the Z scale.

245 is 2.27 standard deviations below the mean

287 is 1.55 standard deviations above the mean

27.211

270245

11270287 .Z

Data = 245 Data = 287

Area = Probability

0.9394 = Area below 287 (1.55)

0.9394 = Area below 287 (1.55)0.0116 = Area below 245 (-2.27)

0.9394 = Area below 287 (1.55)0.0116 = Area below 245 (-2.27)0.9278 = Area between 245 and 287

Example - Solution

0.9278 is the probability that a pregnancy lasts between 245 and 287 days.

We found 245 to be the 1.16 percentile of pregnancy lengths.

We found 287 to be the 93.94 percentile.

Example - Solution

0.9266 is the probability that the woman’s pregnancy had a length that proves the man’s innocence.

Note: This is not the probability the man is not the father. In our way of thinking about things, there is no such thing as “the probability the man is the father.” Either he is or is not the father. (We just don’t know.)

ExampleDetermine the first quartile of the pregnancy lengths.

First draw a picture.

ExampleDetermine the first quartile.

0.2500 = Area

Notice that the Z value will be negative.

0.2500

In problems like these – where the probability is given – the steps of the previous problems are reversed.

ExampleThe steps of the previous problems are reversed.

From the given probability / percentile, determine the Z-score.

Since 0.2500 is the probability of a result less than Q1:

Z -0.67

(Better tools give -0.674.)

0.2500

ExampleYou have the Z score (–0.67). Use either of the given equations to obtain the data value X.

StDevMeanXZ

27067.0

2701167.0 X 27037.7 X

X 37.7270 X63.262

0.2500

Q1 = 262.63

ExampleYou have the Z score (–0.67). Use either of the given equations to obtain the data value X.

StDevZMeanX

1167.0270 X

63.262X

1167.0270 X

37.7270 X

ExampleThe first quartile is 270 – 7.37 = 262.63.(7.37 below the mean)The median is 270.The third quartile is 270 + 7.37 = 277.37.(7.37 above the mean)The IQR is 2(7.37) = 14.74.SD / IQR = 11 / 14.74 = 0.746For any Normal distribution, the SD is about 3/4ths the IQR.

ExampleAt one hospital, women whose pregnancy falls into the top 2% of lengths are recommended for inducing. At what time (in days) is a woman recommended for inducing?

In other words…

Find the length that separates the longest 2% from the shortest 98% of pregnancies.

ExampleFind the length that separates the longest 2% from the shortest 98% of pregnancies.

Area = 0.98

Area = 0.02

Area = 0.98

Area = 0.02X = + Z X = 270 + 2.054 11

Area = 0.98

Area = 0.02X = + Z X = 292.594

ExampleAt one hospital, women whose pregnancy falls into the top 2% of lengths are recommended for inducing. At what time (in days) is a woman recommended for inducing?

ANS: A woman should be induced at 292.6 days.

(This is the 98th percentile of the distribution.)

Percentiles / Z scoresExample

25th percentile Z = -0.67 X = 262.6

Example98th percentile Z = 2.05 X = 292.6

Percentiles / Z scoresExample

25th percentile Z = -0.67 X = 262.6In general for any variable with Normal distribution

The 25th percentile is 0.67 standard deviations below the mean.

Example98th percentile Z = 2.05 X = 292.6

In general for any variable with Normal distributionThe 98th percentile is 2.05 standard deviations above the mean.

IQR Rule of ThumbFor a Normal distribution, ¾ IQR. The standard deviation is fairly close to 75% of the interquartile range.

0.6745 2/3 IQR(Z) (4/3)

IQR (4/3) SD

SD (3/4) IQR

The Normal Models/Distributions

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