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PPT – The Standard Normal Distribution PowerPoint presentation | free to download - id: 790a44-NDM2M

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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 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?

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

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

(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.

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)

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

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

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.

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