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The Standard Normal Distribution

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The Standard Normal Distribution Standard Normal Density Curve The standard normal distribution has a mean = 0 and a standard deviation = 1 Values on the x-axis are ... – PowerPoint PPT presentation

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Title: The Standard Normal Distribution


1
The Standard Normal Distribution
2
Standard Normal Density Curve
  • The standard normal distribution has a mean 0
    and a standard deviation 1
  • Values on the x-axis are the number of standard
    deviations from the mean, called z-scores
  • Interactive Applet www.whfreeman.com/sta
  • What percent of the observations lie within 1
    standard deviation?
  • What if we wanted to know the percent of
    observations that lie within 1.5 standard
    deviations?

3
Using z-score table
  • The area to the left of the corresponding z-
    score is the percentage of observations, P, that
    fall below the z-score.
  • 1. Use the table to find the percentage of values
    below
  • z -1.2.
  • 11.51
  • 2. What is P(z gt -1.2)?
  • 100 11.51 88.49
  • 3. What is P(-1.2 lt z lt 1.2)?
  • 88.49 11.51 76.98

4
Using z-scores with a calculator
  • Use the normalcdf command to compute
    probabilities using the normal curve.
  • Press 2nd, DIST, normalcdf(lower boundary, upper
    boundary). Use -10 or 10 for infinity.
  • For example, to find P(z lt 1) normalcdf(-10, 1)
  • 1. Use the calculator to find the percentage of
    values below
  • z -1.2.
  • 2. What is P(z gt -1.2)?
  • 3. What is P(-1.2 lt z lt 1.2)?

Normalcdf(-10,-1.2) .1151 11.51
Normalcdf(-1.2, 10) .8849 88.49
Normalcdf(-1.2,1.2) .7699 76.99
5
(z- scores)
Z
A positive z- score indicates the number of
standard deviations above the mean A negative z-
score indicates the number of standard deviations
below the mean.
6
Using the normal curve to find percents
  • Find percent below (or the percentile of) a
    certain value
  • Use the formula to find the z-score and use table
    or use calculator normalcdf with lower boundary
    of -10, upper is z
  • Find percent above a certain value
  • Use the formula to find the z-score and subtract
    table value from 100 or use calculator
    normalcdf (lower boundary is z, upper is 10)
  • Find percent between two certain values
  • Use the formula to find two z-scores, one for
    each value
  • Use the table to find two percents and subtract
    them from each other .or.
  • Use calculator normalcdf(lower is 1st z, upper
    is 2nd z)

7
SAT Example
  • The approximately normal distribution of SAT
    scores for the 2009 incoming class at the
    University of Texas had a mean 1815 and a
    standard deviation 252.
  • What is the z- score for a University of Texas
    student who got 1901 on the SAT?
  • Z 1900 1815 0.34
  • 252
  • What percentage did better than this student?
  • P(z gt .34) Normalcdf(.34, 10) .366 or 36.6
  • What percentile was a student
  • at if he made a 2200?
  • z (2200-1815)/252 1.53
  • normalcdf(-10,1.53) 94

8
Heights Example
  • The heights of 18 to 24 year old males in the US
    are approximately normal with mean 70.1 inches
    and standard deviation 2.7 inches. The heights of
    18 to 24 year old females have a mean of 64.8
    inches and a standard deviation of 2.5 inches.
  • Estimate the percentage of US males
  • between 18 and 24 who are 6ft tall or
    taller.
  • Z 72 70.1 .7037
  • 2.7
  • P(z gt .7037) normalcdf(.7037, 10) .2408
    24.08
  • Estimate the percentage of US females between 18
    and 24 who are between 5 feet and 55.
  • Z 60 64.8 -1.92 z 65 64.8 .08
  • 2.5 2.5
  • P(-1.92 lt z lt .08) normalcdf(-1.92,.08) .5044
    50.44

9
Comparison Example
  • Your friend Mark scored a 43 on Mrs. Johnson's
    chemistry test. The mean score in her class was a
    38 and the standard deviation was 4. Marshall
    scored a 67 on his AP Calculus test and the mean
    in that class was a 65, with a standard deviation
    of 7. Whose score was better, relative to their
    classes?
  • Mark z (43 38)/4 1.25
  • P(z lt 1.25) normalcdf(-10, 1.25) .894
    89.4th percentile
  • Marshall z (67 65)/7 .286
  • P(z lt .286) normalcdf(-10, .286) .613
    61.3rd percentile
  • Mark did better because he was
  • further above the mean than Marshall
  • and was in the 89th percentile
  • compared to Marshall who was only
  • in the 61st percentile.

10
Born to Run
  • A study of elite distance runners found a mean
    body
  • weight of 139.1 pounds with a standard deviation
    of 10.6
  • pounds. Assuming the distribution of weights is
  • approximately normal, make a sketch of the weight
    distribution.
  • What percent of runners would have a body
  • weight between 107.3 and 149.7 lb? 84
  • 16 of runners would have a body weight
  • less than how many pounds? 128.5 pounds
  • What percent of runners would have a body weight
    less than 130 pounds? Z (130-139.1)/10.6
    .858 normalcdf(-10, -.858) 19.5
  • What percent of runners would have a body weight
    greater than 160 pounds? Z (160-139.1)/10.6
    1.97 normalcdf(1.97,10) 2.4

107.3 117.9 128.5 139.1 149.7 160.3
170.9
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