# CHAPTER 6: The Standard Deviation as a Ruler & The Normal Model

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

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.

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

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

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.

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?

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

The Empirical Rule

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.

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.

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

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.

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.

Example Problem 2 (p. 123, #18)

Example Problem 2 (p. 123, #18) 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.

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

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.

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.

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

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 X1 = 68 & X2 = 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.

Example Problem 2 (p. 123, #18) d)
P(X > 68 & X< 84) = P(68<X<84) = ? z1 = (68 – 100)/16 = -2.0 z2 = (84 – 100)/16 = -1.0

Example Problem 2 (p. 123, #18) d)
P(X > 68 & X< 84) = P(68<X<84) = P(Z > -2.0 & Z < -1.0) = P(-2.0<Z<-1.0) = ? P(-2.0<Z<-1.0) = P(Z<-1.0) – P(Z<-2.0) =

Example Problem 2 (p. 123, #18) d)
P(-2.0<Z<-1.0) = normalcdf(-2.0,-1.0) = P(Z<-1.0) – P(Z<-2.0) = P(Z<-2.0) = (from the TI-83) P(Z<-1.0) = .0228 P(Z<-1.0) – P(Z<-2.0) = = .1359

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

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

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.

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.

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.

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

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?

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!)

Example Problem 4 (p. 125, #36) Area to the left = 1.0 - .05 = .95
A) the highest 5% of all IQ’s? Area to the left = = .95 z = invNorm(.95) = 1.645 X = z*σ + μ = 1.645* = Approximately 5% of people have an IQ greater than

Example Problem 4 (p. 125, #36) Area to the left = .30
B) the lowest 30% of the IQ’s? Area to the left = .30 z = invNorm(.30) = X = z*σ + μ = * = 91.61 Approximately 30% of people have an IQ lower than

Example Problem 4 (p. 125, #36) Area to the left of X1 = .10
C) the middle 80% of the IQ’s? Area to the left of X1 = .10 z = invNorm(.10) = -1.28 X = z*σ + μ = -1.28* = 79.50

Example Problem 4 (p. 125, #36) Area to the left of X2 = .90
C) the middle 80% of the IQ’s? Area to the left of X2 = .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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