Exercise: Empirical Rule

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Use the empirical rule to answer the following:

! Monthly maintenance costs are distributed normally with a µ=$250 and σ=$50 " 1) What percent of months have maintenance

costs in the range of $200 to $300?

" 2) What is the chance (i.e. probability) that a randomly chosen month has a maintenance cost of $150 or less?

Answer:

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" 1)

" 2)

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5.2 Properties of the Normal Distribution part 2

! Standard Scores

! Finding percentiles when you can’t use the empirical rule (when the data value is something other than 1, 2 or 3 standard deviations from the mean)

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! 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

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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 (continued): 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 IQ of 85: z = = -0.94

85 – 100 16

standard score for IQ of 100: z = = 0.00

100 – 100 16

standard score for IQ of 125: z = = 1.56

125 – 100 16

An IQ of 85 is 0.94 standard deviations below the mean.

Standard Scores ! What percentile are the following IQ scores?

"  85 (standard score: z = -0.94)

" 100 (standard score: z = 0)

" 125 (standard score: z = 1.56)

! We can’t use the empirical rule here.

! We’ll have to use a table to find the percentages (Appendix A in our book).

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A subset of Appendix A (shown here) is provided in Section 5.2 in the book. This table shows the percentage of observations below any given standard score.

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What percent of observations are below a standard score of z = -0.94? The closest standard score is -0.95 and 17.11% of the observations are below a standard score of -0.95. 17% shaded

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What percent of observations are below a standard score of z = 0? Ans: 50% A standard score of 0 is at the 50th percentile.

50% shaded

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What percent of observations are below a standard score of z = 1.56? This standard score lies between 1.5 and 1.6 on the table. We can approximate the percentile for this standard score as (93.32% + 94.52%) 2 Or 93.92%, which is the 93.92nd percentile.

94% shaded

Standard Scores ! What percentile are the following IQ scores?

" IQ of 85 (z=-0.94) is at the 17th percentile. " IQ of 100 (z=0) is at the 50th percentile " IQ of 125 (z=1.56) is at the 94th percentile.

" Recall that negative z-scores are below the mean and positive z-scores are above the mean.

! Thus, we can get the percentiles even though we’re not exactly 1, 2 or 3 standard deviations from the mean. 12

! We will use Appendix A from the book (a subset of that table was shown above) to compute percentiles and probabilities because it has finer resolution (more decimals).

! THIS TAKES PRACTICE!

! Active work: See worksheet on normal curve scores.

Standard Scores

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More Exercises: z-scores

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! Assume you have a normal distribution. Use the z-score table in Appendix A to answer: " 1) What percent of observations lie below a z-

score of 0?

" 2) What percent of observations lie below a z-score of 1.72?

More Exercises: z-scores

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" 3) What percent of observations fall BETWEEN z-scores of 0 and 1.72?

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5.2 Properties of the Normal Distribution part 3

! Connecting z-scores to probabilities.

! Example: The Stanford-Binet IQ test is normally distributed and scaled so that scores have a mean of 100 and a standard deviation of 16.

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! Example: The Stanford-Binet IQ test is normally distributed and scaled so that scores have a mean of 100 and a standard deviation of 16.

" If you draw someone at random, what is the probability that they have an IQ score of 90 or less?

" To answer this, we just need to know what percent of IQ scores are at 90 or lower.

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! Example: The Stanford-Binet IQ test is normally distributed with a mean of 100 and standard deviation of 16.

Let X be an IQ score of a person.

Short-hand notation: X ~ N(µ=100,σ=16)

19 Normal

The 2 parameters needed to define a normal distribution.

“is distributed”

" If you draw someone at random, what is the probability that they have an IQ score of 90 or less?

" We need to answer:

When X ~ N(µ=100,σ=16), what is P(X ≤ 90)? X is a data value (or IQ score in this case). We will convert it to a z-score…

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" P(X ≤ 90) = P( ≤ )

= P(Z ≤ – 10/16)

= P(Z ≤ – 0.63)

= 0.2643 21

data value – mean standard deviation z = standard score =

X –µ σ

90 –100 16

" P(X ≤ 90) = P(Z ≤ – 0.63) = 0.2643

" The probability of randomly drawing someone with an IQ score of 90 or lower is 0.2643.

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data value – mean standard deviation z = standard score =

An IQ score of 90 has a z-score of - 0.63

Looked up on z-table

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QUICK-CHECK: The Empirical Rule tells me the percent that is below an IQ of 90 has to be between16% (to the left of 1σ below the mean) and 50% (to the left of the mean itself).

IQ 90

So, 26.43% is totally in-line with my Empirical Rule information because being 0.63 standard deviations is between 1 and 0 standard deviations down from the mean.

Using new notation, exercise 1: !  Let X ~ N(µ=40,σ=5). Find P(X < 51):

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Using new notation, exercise 2: ! Suppose bowling scores are normally

distributed with a mean of 186 and a standard deviation of 30. Find the percentage of games with a score of 120 or HIGHER.

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