1. 4/29/13 Have out: Bellwork: assignment, graphing calculator,
pencil, red pen, highlighter
Have out:
Bellwork:
If the mean test score was a 75 with a standard deviation of 10.5, calculate the z–scores for the test scores in parts (a) – (c). Also, determine if they are outliers.
+ 1
a) 23
b) 100
c) 80
+ 1
d) Victoria has a z–score of – What is her raw score. Round!
+ 1
+ 1
+ 1
+ 1
+ 1
outlier!
outlier!
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+ 1
total:
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2. Percentile Percentile percentage
____________: a ranking that indicates the ___________
of students who scored _______ that score.
below
Example: Consider the following Geometry test scores sorted in
ascending order:
{17, 42, 53, 56, 57, 58, 60, 60, 69, 71, 72, 75, 75, 75, 76, 79, 80,
81, 83, 86, 86, 89, 89, 90, 92, 93, 93, 100}
a) What is Ryan’s percentile if he scored an 80?
b) What is Beatriz’s percentile if she scored a 71? 16 9
57%
32% 28 28
c) What is Harry’s percentile if he scored a 53?
d) What is Emily’s percentile if she scored a 92? 2 24 7%
86% 28 28

3. The Normal Distribution
If you toss a coin 8 times, then theoretically, what is the number of
possible outcomes for the number of heads? We can just look at
the 8th row of Pascal’s triangle: x 1 2 3 4 5 6 7 8
P(X)

4. x 1 2 3 4 5 6 7 8
P(X)
Make a relative-frequency histogram illustrating the number of heads and the frequency of each type.
P(x)
Connect the points. Notice that it forms a ____ - _______ ______.
bell
shaped
curve
Frequency x 1 2 3 4 5 6 7 8
# of heads

5. A distribution such as the one above is called a _________________.
P(x)
A distribution such as the one above is called a _________________.
normal distribution
Frequency x 1 2 3 4 5 6 7 8
# of heads
This example is a ________ probability distribution because there is only a countable number of possible values.
discrete
On the other hand, a __________ probability distribution can be any value in an interval of real numbers. Continuous probability distributions are represented by ______, not histograms.
continuous
curves
The_____ and ______________ determine the shape of the curve.
mean
standard deviation

6. On this normal curve below, place: , , , and .
68%
95%
99.7%
For a normal distribution, 1 68
____ % of the data falls within ___ σ of
____ % of the data falls within ___ σ of 95 2
99.7 3
____ % of the data falls within ___ σ of
This is called the __________ ______.
Empirical
Rule
(by experience or results in a population)
Normal curve animation

7. 47.5% - 34% = 13.5%
50%
50% - 34% = 16%
50% % = 2.5%
34%
68%
95%
99.7%
Normal curve animation

8. a) What percent of men have heights over 69 inches?
Example # 1: The heights of adult American men are normally distributed with a mean of 69 inches and a standard deviation of 2.5 inches. Sketch a normal curve, and label , , , and
a) What percent of men have heights over 69 inches?
b) What percent of men are taller than 71.5 inches?
50%
50% – 34%
= 16%
percent
σ = 2.5
34% //
61.5 64
66.5 69
71.5 74
76.5
Height –3 –2 –1 1 2 3
z-score

9. c) What percent of men are shorter than 64 inches?
50% – 13.5% – 34%
= 2.5%
d) What percent of men are between 64 and 71.5 inches?
13.5% + 68%
= 81.5%
percent
σ = 2.5
34%
68%
13.5% //
61.5 64
66.5 69
71.5 74
76.5
Height –3 –2 –1 1 2 3
z-score

10. a) What is the percentile for a score of 110?
Example # 2: The Stanford – Binet IQ Test is normally distributed with a mean of 100 and standard deviation of 10. Sketch a normal curve, and label , , , and
a) What is the percentile for a score of 110?
50% + 34%
= 84%
b) What percent of people score over 120?
50% – 34% – 13.5%
= 2.5%
Percent
34% %
σ = 10
50%
34% //
Score 70 80 90
100
110
120
130 –3 –2 –1 1 2 3
z-score

11. c) What is the percent for a score of 80?
50% %
= 97.5%
d) What percent of people score between 80 and 110?
68% %
= 81.5%
Percent
50%
47.5%
σ = 10
68%
13.5% //
Score 70 80 90
100
110
120
130 –3 –2 –1 1 2 3
z-score

12. e) Forest Gump’s score was 76. What was his z–score?
f) Is Forest Gump an outlier?
Yes, his score is an outlier because the z-score is more than 2σ below the mean.
Percent
σ = 10 //
Score 70 80 90
100
110
120
130 –3 –2 –1 1 2 3
z-score

13. Example # 3: The heights of American women, aged 18 – 24, are normally distributed, with a mean of 64.5 inches and a standard deviation of 2.5 inches. Sketch a normal curve, and label , , , and
In a random sample of 2000 young women, how many would you expect to be:
a) Over 64.5 inches?
b) Under 62 inches?
50%
50% – 34%
= 16%
2000(0.5)
= 1000 women
2000(0.16)
= 320 women
Percent
σ = 2.5 // 57
59.5 62
64.5 67
69.5 72
Height –3 –2 –1 1 2 3
z–score

14. c) Between 62 and 64.5 inches? 34% 2000(0.34) = 680 women
d) Under 67 inches?
50% + 34%
= 84%
2000(0.84)
=1680 women
Percent
σ = 2.5
50%
34%
34% // 57
59.5 62
64.5 67
69.5 72
Height –3 –2 –1 1 2 3
z–score

15. e) Over 69.5 inches? 50% – 34% – 13.5% = 2.5% 2000(0.025) = 50 women
f) Under 57 inches?
50% – 49.85%
= 0.15%
2000(.0015)
= 3 women
Percent
σ = 2.5
13.5%
= 49.85%
34% // 57
59.5 62
64.5 67
69.5 72
Height –3 –2 –1 1 2 3
z–score

16. Finish the worksheet.

17. The_____ and ______________ determine the shape of the curve. mean
standard deviation
Practice # 1: On the axes provided, sketch 2 different normal curves with the same mean but different standard deviations.
Y =
\ Y1 =
2nd
DISTR
1: normalpdf (
ENTER
X, T, , n , , 1 ) y x
\ Y2 =
2nd
DISTR
1: normalpdf (
ENTER
X, T, , n , ,
1.5 ) x σ
y2 is lower because the total area under the curve is still 1 (100%). y1
Both have the same amount of data. y2
Normal curve animation