STAT1010 – Empirical rule

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Review Exercises: Recall…

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Given these 3 things: P(R)=0.10

P(H)=0.20

P(H|R)=0.20

low

high

Find: 1) P(R) = P(not R)=

2) P(H|R)=P(not H | R)=

(mosaic plot) 1.00

0.75

0.50

0.25

0.00

Review Exercises:

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3) Suppose P(A)=0.5, P(B)=0.75, P(A and B)=0.4

Find: P(A or B)

4) Suppose P(A)=0.2, P(B)=0.65, P(A or B)=0.75

Find: P(A and B)

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5.2 Properties of the Normal Distribution

! READ THIS SECTION IN THE BOOK!!! " Understand the examples.

STAT1010 – Empirical rule

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5.2 Properties of the Normal Distribution

!  Important

!  Important

!  Important

Normal Distribution ! Most widely used distribution. ! Arises naturally in physical phenomena ! Two parameters COMPLETELY define a

normal distribution, µ and σ. !  µ is the center (mean) of the distribution. ! σ is the standard deviation of the distribution

(quantifies spread). ! Symmetrical distribution.

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Normal Distribution ! Can occur anywhere along the real number

line. µ specifies the center (position) and σ specifies the spread.

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Different normal distributions for selected values of the parameters µ and σ. Recall that the area under the curve is 1 (or 100%)

STAT1010 – Empirical rule

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Empirical Rule (68-95-99.7 Rule)

! Special result of the normal distribution:

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Empirical Rule (68-95-99.7 Rule)

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100% of the data falls in (- , )

Consider the normal model with µ = 0 and σ =1.

µ

∞ ∞

Empirical Rule (68-95-99.7 Rule)

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!  68% of the values fall within 1 standard deviation of the mean.

1σ

68%

STAT1010 – Empirical rule

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Empirical Rule (68-95-99.7 Rule)

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!  95% of the values fall within 2 standard deviations of the mean.

2σ

95%

Empirical Rule (68-95-99.7 Rule)

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!  99.7% of the values fall within 3 standard deviations of the mean.

3σ

99.7% 0.15% left in this tail 0.15% left in this tail

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Empirical Rule (68-95-99.7 Rule)

!  Very useful for determining what % of the observations fall between two x-values

!  Very useful for determining what % of the observations fall in the tail

STAT1010 – Empirical rule

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Finding a Percentile using the Empirical Rule

! Example 1: The amount of cereal in a box varies a little from box to box. Suppose the amount in a box has a normal distribution with µ=15 oz. and σ=0.2 oz.

What percentage of boxes have between 14.6 and 15.4 oz. of cereal?

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Finding a Percentile using the Empirical Rule

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14.6 14.8 15 15.2 15.4 ß Specific to cereal

What percent is within 2 standard deviations of the mean?

Finding a Percentile using the Empirical Rule

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Example 1: What percentage of boxes have between 14.6 and 15.4 oz. of cereal? Two standard deviations down from µ:

15 – 2(0.2) = 14.6 Two standard deviations up from µ:

15 + 2(0.2) = 15.4

Answer: 95%

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Finding a Percentile using the Empirical Rule

! Example 2: Octane ratings are normally distributed with µ=91 and σ=1.5

What percentage of octane ratings fall below 92.5?

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Finding a Percentile using the Empirical Rule

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68% of the values fall between 89.5 and 92.5 95% of the values fall between 88.0 and 94.0 99.7% of the values fall between 86.5 and 95.5

(µ - 1σ, µ + 1σ)

(µ - 2σ, µ + 2σ)

(µ - 3σ, µ + 3σ)

Finding a Percentile using the Empirical Rule

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! What percentage of octane ratings fall below 92.5?

! Draw a picture. ! How far away from the mean is 92.5 in

terms of number of standard deviations?

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Draw a picture

Finding a Percentile using the Empirical Rule

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! What percentage of octane ratings fall below 95.5?

! Draw a picture. ! How far away from the mean is 95.5 in

terms of number of standard deviations?

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Draw a picture

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Finding a Percentile using the Empirical Rule

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! Example 3: On a visit to the doctor’s office, your fourth-grade daughter is told that her height is 1 standard deviation above the mean for her age and sex. What is her percentile for height? Assume that heights of fourth-grade girls are normally distributed.

Finding a Percentile using the Empirical Rule

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The total percent in green is 84% (this represents percentage of heights below 1 standard deviation above the mean). The girl’s height is at the 84th percentile.

Identifying Unusual Results

! For a normal distribution, 95% of all observations lie within 2 standard deviations of the mean.

! As a rule of thumb, “unusual values” are values that are more than 2 standard deviations away from the mean.

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STAT1010 – Empirical rule

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! What about percentages that are not exactly 68% or 95% or 99.7%?

! What if we’re 1.5 standard deviations up from the mean? How do we compute such a percentile?

! Solution: Standard Scores

Normal Percentiles

Standard Scores

!  Language like “one standard deviation from the mean” is very generic and can be applied to ANY normal distribution (i.e. any µ and any σ).

! This makes using the empirical rule very useful whether you’re dealing with bowling scores, weights, heights, etc.

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Standard Scores

! When we can’t use the empirical rule, we will instead relate our specific normal distribution (i.e. a particular µ and σ) to a very special normal distribution called the ‘Standard Normal’ distribution which has µ=0 and σ=1.

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STAT1010 – Empirical rule

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Standard Scores ! Every normal distribution can be related to

the standard normal (µ=0,σ=1).

! To do this, we quantify “how many standard deviations from the mean” any particular value is.

! The number of standard deviations a data value is above or below the mean is called it’s standard score (or z-score).

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Standard Scores

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The number of standard deviations a data value lies above or below the mean is called its standard score (or z-score), defined by z = standard score =

The standard score is positive for data values above the mean and negative for data values below the mean.

data value – mean standard deviation

Standard Scores ! Example 1: For our cereal, the amount in

a box has a normal distribution with µ=15 oz. and σ=0.2 oz.

! How many standard deviations away from the mean is a box with 15.25 oz?

! What is the standard score of a box with 15.25 oz?

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STAT1010 – Empirical rule

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Standard Scores

!  15.25 is 0.25 oz. up from the mean. ! How many standard deviations up from the

mean? " Recall, 1 standard deviation is 0.2 oz

Compute: 0.25/0.2=1.25 The distance 0.25 is 1.25 standard deviations. So, 15.25 is 1.25 standard deviations up from the mean.

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15 + (1.25 x 0.2)=15.25

mean 1.25 standard deviations

z = standard score = = = 1.25 data value – mean standard deviation

15.25 - 15 0.20

Standard Scores

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15.25 - 15 0.2

‘Standard Normal’ distribution (µ=0,σ=1)

! The standard score for a box with 15.25 oz is

z = standard score = = 1.25 (this z-score is positive because it is above the mean)

Standard Scores ! Example 2: The Stanford-Binet IQ test is

scaled so that scores have a mean of 100 and a standard deviation of 16. Find the standard scores for IQs of 85, 100, and 125.

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standard score for 85: z = = -0.94

85 – 100 16

standard score for 100: z = = 0.00

100 – 100 16

standard score for 125: z = = 1.56

125 – 100 16

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Standard Scores ! We can interpret these standard scores as

follows: " 85 is 0.94 standard deviation below the mean,

100 is equal to the mean, and 125 is 1.56 standard deviations above the mean.

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Standard Scores ! The Stanford-Binet IQ test.

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