The Standard Deviation as a Ruler and the

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Chapter 6

• The Standard Deviation as a Ruler and the
• Normal Model

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The Standard Deviation as a Ruler

• A student got a 67/75 on the first exam and a
  64/75 on the second exam. She was disappointed
  that she did not score as well on the second
  exam.
• To her surprise, the professor said she actually
  did better on the second exam, relative to the
  rest of the class.

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The Standard Deviation as a Ruler

• How can this be?
• Both exams exhibit variation in the scores.
• However, that variation may be different from one
  exam to the next.
• The standard deviation provides a ruler for
  comparing the two exam scores.

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Summarizing Exam Scores

• Exam 1
• Mean
• Standard
• Deviation

• Exam 2
• Mean
• Standard
• Deviation

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Standardizing Variables

• z has no units (just a number)
• Puts variables on same scale
• Center (mean) at 0
• Spread (standard deviation) of 1
• Does not change shape of distribution

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Standardized Exam Scores

• Exam 2
• Score 64

• Exam 1
• Score 67

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Standardized Exam Scores

• On exam 1, the 67 was 0.87 standard deviations
  better than the mean.
• On exam 2, the 64 was 1.17 standard deviations
  better than the mean.

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Standardizing Variables

• z of standard deviations away from mean
• Negative z number is below mean
• Positive z number is above mean

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Standardizing Variables

• Height of women
• Height of men
• Jill is 69 inches tall
• Jack is 72 inches tall
• Who is taller (comparatively)?

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Standardizing Variables
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Standardizing Variables

• Jill is 1.2 standard deviations above mean height
  for women
• Jack is 0.67 standard deviations above mean
  height for men
• Jill is taller (comparatively)

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Normal Distributions

• Bell Curve
• Physical Characteristics
• Examples
• Heights
• Weights
• Lengths of Bird Wings
• Most important distribution in this class.

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Normal Distributions

• Curve is always above the x-axis
• The tails never reach the x-axis
• They continue on to infinity and infinity
• Area under entire curve 1 or 100
• Mean Median
• This means the curve is symmetric

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Normal Distributions

• Two parameters (not calculated, i.e. do not come
  from the data)
• Mean µ (pronounced meeoo)
• Locates center of curve
• Splits curve in half
• Shifts curve along x-axis

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Normal Distributions

• Standard deviation s (pronounced sigma)
• Controls spread of curve
• Smaller s makes graph tall and skinny
• Larger s makes graph flat and wide
• Ruler of distribution
• Write as N(µ,s)

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Standard Normal Distribution

• Puts all normal distributions on same scale
• z has center (mean) at 0
• z has spread (standard deviation) of 1

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Standard Normal Distribution

• z of standard deviations away from mean µ
• Negative z, number is below the mean
• Positive z, number is above the mean
• Written as N(0,1)

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Standardizing

• Y N(70,3). Standardize y 68.
• y 68 is 0.67 standard deviations below the mean

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Standardizing

• Notice the difference between Y and y
• Y denotes an entire distribution all possible
  values of the distribution, the shape, the
  center, the spread
• y denotes a single value ONLY
• Can be generalized to all capital letters
• Z and z
• X and x

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Standardizing
N(70,3)
N(0,1)
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Standardizing

• Y N(70,3). Standardize y 74
• y 74 is 1.33 standard deviations above mean

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Standardizing
N(70,3)
N(0,1)
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Standardizing

• Y N(65,2.5). Standardize y 63
• Y N(65,2.5). Standardize y 68

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Standardizing

• The problem with standardizing is that it only
  tells me where the value is on the curve
• Probabilities are areas under the curve
• P(Y 68) 0 (ALWAYS)
• However, P(Y gt 68) or P(Y lt 68) gt 0
• How do we find these probabilities?

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68-95-99.7 Rule

• 68 of observations are within 1 s of the mean µ
• For N(0,1) this is between 1 and 1

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68-95-99.7 Rule

• 95 of observations are within 2 s of the mean µ
• For N(0,1) this is between 2 and 2

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68-95-99.7 Rule

• 99.7 of observations are within 3 s of the mean
  µ
• For N(0,1) this is between 3 and 3

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68-95-99.7 Rule

• Given Y (the heights of men) is N(70,3), between
  what two values are 68 of the data?
• 68 of the data are between the values 67 and 73
• Between what two numbers are 95 of the data?
• 95 of the data are between the values 64 and 76
• Between what two numbers are 99.7 of the data?
• 99.7 of the data are between the values 61 and 79

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68-95-99.7 Rule

• P(Y lt 70)?
• 0.5
• P(Y lt 67)?
• (1 0.68)/2 0.16
• P(Y lt 64)?
• (1 0.95)/2 0.025
• P(Y lt 61)?
• (1 0.997)/2 0.0015

• P(Y gt 67)?
• 0.5 (0.68)/2 0.84
• P(Y gt 64)?
• 0.5 (0.95)/2 0.975
• P(Y gt 61)?
• 0.5 (0.997)/2 0.9985

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68-95-99.7 Rule

• P(Y gt 70)?
• 0.5
• P(Y gt 73)?
• (1 0.68)/2 0.16
• P(Y gt 76)?
• (1 0.95)/2 0.025
• P(Y gt 79)?
• (1 0.997)/2 0.0015

• P(Y lt 73)?
• 0.5 (0.68)/2 0.84
• P(Y lt 76)?
• 0.5 (0.95)/2 0.975
• P(Y lt 79)?
• 0.5 (0.997)/2 0.9985

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68-95-99.7 Rule

• P(67 lt Y lt 70)?
• 0.68/2 0.34
• P(67 lt Y lt 76)?
• (0.68/2) (0.95/2) 0.815
• P(67 lt Y lt 79)?
• (0.68)/2 (0.997)/2 0.8385
• P(64 lt Y lt 79)?
• (0.95)/2 (0.997)/2 0.9735

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Areas under curve

• Another way to find probabilities when values are
  not exactly 1, 2, or 3 ? away from µ is by using
  the Normal Values Table
• Gives amount of curve below a particular value of
  z
• z values range from 3.99 to 3.99
• Row ones and tenths place for z
• Column hundredths place for z

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Finding Values

• What percent of a standard Normal curve is found
  in the region Z lt -1.50?
• P(Z lt 1.50)
• Find row 1.5
• Find column .00
• Value 0.0668

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Finding Values

• P(Z lt 1.98)
• Find row 1.9
• Find column .08
• Value 0.9761

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Finding values

• What percent of a std. Normal curve is found in
  the region Z gt-1.65?
• P(Z gt -1.65)
• Find row 1.6
• Find column .05
• Value from table 0.0495
• P(Z gt -1.65) 0.9505

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Finding values

• P(Z gt 0.73)
• Find row 0.7
• Find column .03
• Value from table 0.7673
• P(Z gt 0.73) 0.2327

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Finding values

• What percent of a std. Normal curve is found in
  the region 0.5 lt Z lt 1.4?
• P(0.5 lt Z lt 1.4)
• Table value 1.4 0.9192
• Table value 0.5 0.6915
• P(0.5 lt Z lt 1.4) 0.9192 0.6915 0.2277

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Finding values

• P(2.3 lt Z lt 0.05)
• Table value 0.05 0.4801
• Table value 2.3 0.0107
• P(2.3 lt Z lt 0.05) 0.4801 0.0107
  0.4694

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Finding values

• Above what z-value do the top 15 of all z-value
  lie, i.e. what value of z cuts off the highest
  15?
• P(Z gt ?) 0.15
• P(Z lt ?) 0.85
• z 1.04

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Finding values

• Between what two z-values do the middle 80 of
  the obs lie, i.e. what values cut off the middle
  80?
• Find P(Z lt ?) 0.10
• Find P(Z lt ?) 0.90
• Must look inside the table
• P(Zlt-1.28) 0.10
• P(Zlt1.28) 0.90

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Solving Problems

• The height of men is known to be normally
  distributed with mean 70 and standard deviation
  3.
• Y N(70,3)

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Solving Problems

• What percent of men are shorter than 66 inches?
• P(Y lt 66) P(Zlt ) P(Zlt-1.33)
  0.0918

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Solving Problems

• What percent of men are taller than 74 inches?
• P(Y gt 74) 1-P(Ylt74) 1 P(Zlt )
  1 P(Zlt1.33) 1 0.9082 0.0918

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Solving Problems

• What percent of men are between 68 and 71 inches
  tall?
• P(68 lt Y lt 71) P(Ylt71) P(Ylt68)P(Zlt
  )-P(Zlt )P(Zlt0.33) - P(Zlt-0.67) 0.6293
  0.2514 0.3779

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Solving Problems

• Scores on SAT verbal are known to be normally
  distributed with mean 500 and standard deviation
  100.
• X N(500,100)

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Solving Problems

• Your score was 650 on the SAT verbal test. What
  percentage of people scored better?
• P(X gt 650) 1 P(Xlt650) 1 P(Zlt
  ) 1 P(Zlt1.5) 1 0.9332
  0.0668

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Solving Problems

• To solve a problem where you are looking for
  y-values, you need to rearrange the standardizing
  formula

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Solving Problems

• What would you have to score to be in the top 5
  of people taking the SAT verbal?
• P(X gt ?) 0.05?
• P(X lt ?) 0.95?

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Solving Problems

• P(Z lt ?) 0.95?
• z 1.645
• x is 1.645 standard deviations above mean
• x is 1.645(100) 164.5 points above mean
• x 500 164.5 664.5
• SAT verbal score at least 670

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Solving Problems

• Between what two scores would the middle 50 of
  people taking the SAT verbal be?
• P(x1 ? lt X lt x2?) 0.50?
• P(-0.67 lt Z lt 0.67) 0.50
• x1 (-0.67)(100)500 433
• x2 (0.67)(100)500 567

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Solving Problems

• Cereal boxes are labeled 16 oz. The boxes are
  filled by a machine. The amount the machine
  fills is normally distributed with mean 16.3 oz
  and standard deviation 0.2 oz.

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Solving Problems

• What is the probability a box of cereal is
  underfilled?
• Underfilling means having less than 16 oz.
• P(Y lt 16) P(Zlt ) P(Zlt -1.5)
  0.0668

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Solving Problems

• A consumer group wants to the company to change
  the mean amount of cereal the machine fills so
  that only 3 of boxes are underfilled. What do
  we need to change the mean to?
• P(Y lt 16) 0.03
• What is z so that P(Z lt ?) 0.03?
• z 1.88

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Solving Problems

• 16 must be 1.88 standard deviations below mean.
• 16 must be 1.88(0.2) 0.376 below mean
• Mean 16 0.376 16.376

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Solving Problems

• Company president feels that is too much cereal
  to put in each box. She wants to set the mean
  weight on the machine to 16.2, but only have 3
  of the boxes underfilled.
• How can she do this?
• Change the standard deviation of the machine.

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Solving Problems

• P(Y lt 16) 0.03
• What is z so that P(Z lt ?) 0.03?
• z 1.88
• 16 must be 1.88 standard deviations below 16.2
• 0.2 1.88 s
• s 0.106

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Are Your Data Normal

• The histogram should be mounded in the middle and
  symmetric.
• The data plotted on a normal probability
  (quantile) plot should follow a diagonal line.

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Are Your Data Normal?
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Are Your Data Normal
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Check Normal Assumption

• Let W be the price of 1 1/2 and 2 story houses
  sold in Ames between 9-2004 and 10-2005
• We are told that 204,500 and that
  92,350. We decide to model the price of homes
  with a normal model with out plotting any data.
  Thus we assume that
• We want to find the percent of homes that sold
  for more than 350,000 or P(W gt 350,000).
• We find that
• We then use the table to find that
  P(Wgt200,000).519 or 51.9