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CHAPTER 6: The Standard Deviation as a Ruler & The Normal Model KENNESAW STATE UNIVERSITY MATH 1107

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MOTIVATING EXAMPLE: Most high school students in the U.S. take either the SAT or ACT in preparation to apply to college. Both exams are designed to measure high school academic achievement.

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MOTIVATING EXAMPLE: SAT (Verbal & Math proportions combined) – Total scores range from 400 to 1600 – The national mean is around 1020 – The national standard deviation is around 110

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MOTIVATING EXAMPLE: ACT – Total scores range from 1 to 36 – The national mean is around 21.0 – The national standard deviation is around 4.7

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MOTIVATING EXAMPLE: You have studied, prepared, and taken the SAT (and done quite well on it = 1305). HOWEVER, one of the main schools you want to apply to ONLY accepts the ACT.

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MOTIVATING EXAMPLE: We can’t compare SAT & ACT scores directly because they use different scales: – This would be like comparing apples & oranges. But can we convert your SAT score to an ACT score? How?

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MOTIVATING EXAMPLE: But can we convert your SAT score to an ACT score? YES!!! How? Z-scores!!!

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The Empirical Rule

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Example Problem 1 (p. 122, #5) A town’s January high temperatures average 36 degrees F with a standard deviation of 10 degrees, while in July the mean high temperature is 74 degrees and the standard deviation is 8 degrees. In which month is it more unusual to have a day with a high temperature of 55 degrees? Explain.

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Example Problem 1 (p. 122, #5) Strategy: Find a z-score for X = 55 degrees for both months. Then compare the z-scores. Which ever one has a more extreme value (a larger absolute value) will be the more unusual occurrence.

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Example Problem 1 (p. 122, #5) For January: z Jan = (55 – 36)/10 = 1.9 For July: z Jul = (55 – 74)/8 =

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Example Problem 1 (p. 122, #5) CONCLUSION: It is more unusual to have a day with a high temperature of 55 degrees F in July: A high of 55 degrees is standard deviations below the mean temperature for July. In contrast, a high of 55 degrees is only 1.9 standard deviations above the mean for January.

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Example Problem 2 (p. 123, #18) IQ. Some IQ tests are standardized to a Normal model with a mean of 100 and a standard deviation of 16. – A) Draw the model for these IQ scores. Clearly label it, show what the Empirical Rule predicts about the scores.

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Example Problem 2 (p. 123, #18)

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b) In what interval would you expect the central 95% of IQ scores to be found? – Within 2 standard deviations of the mean, therefore between 68 & 132 IQ points.

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Example Problem 2 (p. 123, #18) c) About what percent of people should have IQ scores above 116?

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Example Problem 2 (p. 123, #18) Strategy: – Define the random variable of interest. – State the question in appropriate statistical notation. – Use the picture you have already drawn to shade the appropriate region. – Find a z-score for X = 116. – Find the appropriate area to the left or right of the z- score. Summarize the results in a complete sentence.

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Example Problem 2 (p. 123, #18) c) X = IQ score of interest P(X > 116) = ? z = (116 – 100)/16 = 1.0 P(z > 1.0) = P(X > 116) =.1587 ANSWER: We expect approximately 15.87% of people to have an IQ above 116.

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Example Problem 2 (p. 123, #18) d) About what percent of people should have IQ scores between 68 and 84?

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Example Problem 2 (p. 123, #18) Strategy: – Define the random variable of interest. (Already done) – State the question in appropriate statistical notation. – Use the picture you have already drawn to shade the appropriate region. – Find z-scores for both X 1 = 68 & X 2 = 84. – Find the appropriate area to the left of each z-score separately. Then take the difference between them. Summarize the results in a complete sentence.

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Example Problem 2 (p. 123, #18) d) P(X > 68 & X< 84) = P(68

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Example Problem 2 (p. 123, #18) d) P(X > 68 & X< 84) = P(68

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Example Problem 2 (p. 123, #18) d) P(-2.0

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Example Problem 2 (p. 123, #18) d) Approximately 13.59% of people possess an IQ score between 68 and 84.

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Example Problem 2 (p. 123, #18) e) About what percent of people should have IQ scores above 132?

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Example Problem 2 (p. 123, #18) Strategy (The same as before!!!!): – Define the random variable of interest. (Already done) – State the question in appropriate statistical notation. – Use the picture you have already drawn to shade the appropriate region. – Find a z-score for X = 132. – Find the appropriate area to the left or right of the z- score. Summarize the results in a complete sentence.

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Example Problem 2 (p. 123, #18) e) P(X > 132) = ? z = (132 – 100)/16 = 2.0 P(X > 132) = P(Z > 2.0) =.0228 Approximately 2.28% of people have an IQ score above 132.

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Example Problem 3 (taken from p. 123, #23) The diameter of trees in a forest for distributed N(10.4, 4.7). a) Draw the Normal model for these tree diameters.

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Example Problem 3 (taken from p. 123, #23)

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Example Problem 4 (p. 125, #36) More IQ. In the Normal model N(100, 16) what cutoff value bounds. – A) the highest 5% of all IQ’s? – B) the lowest 30% of the IQ’s? – C) the middle 80% of the IQ’s?

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Example Problem 4 (p. 125, #36) – A) the highest 5% of all IQ’s? Draw a picture!! Shade the region of interest Find the area to the left of the cutoff value for the region of interest. Use the invNorm function on the TI-83 to obtain the corresponding z-score. Transform that z-score to a X (an IQ score!)

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Example Problem 4 (p. 125, #36) – A) the highest 5% of all IQ’s? Area to the left = =.95 z = invNorm(.95) = X = z*σ + μ = 1.645* = Approximately 5% of people have an IQ greater than

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Example Problem 4 (p. 125, #36) B) the lowest 30% of the IQ’s? Area to the left =.30 z = invNorm(.30) = X = z*σ + μ = * = Approximately 30% of people have an IQ lower than

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Example Problem 4 (p. 125, #36) C) the middle 80% of the IQ’s? Area to the left of X 1 =.10 z = invNorm(.10) = X = z*σ + μ = -1.28* = 79.50

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Example Problem 4 (p. 125, #36) C) the middle 80% of the IQ’s? Area to the left of X 2 =.90 z = invNorm(.90) = 1.28 X = z*σ + μ = 1.28* = Approximately 80% of the middle IQ scores fall between and

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