Introduction to the 3 different "meanings" of "area under the normal curve" and the ShadeNorm() command on the TI graphing calculators.

1. Working with z-scores
and the Normal Curve
... it's everywhere!
The Bell Curve by flickr user c r i s

2. ... but first a little quiz ... ready?
Studying by scui3asteveo

3. HOMEWORK
The table shows the lengths in millimetres of 52 arrowheads.
16 16 17 17 18 18 18 18 19 20 20 21 21
21 22 22 22 23 23 23 24 24 25 25 25 26
26 26 26 27 27 27 27 27 28 28 28 28 29
30 30 30 30 30 30 31 33 33 34 35 39 40
(b) Determine the lengths of arrowheads
(a) Calculate the mean length
one standard deviation below and one
and the standard deviation.
standard deviation above the mean.
(d) What percent of the arrowheads
(c) How many arrowheads
are within one standard deviation
are within one standard
of the mean length?
deviation of the mean?

4. North American women have a mean height of 161.5 cm and a
standard deviation of 6.3 cm.
(a) A car designer designs car seats to fit women taller than
159.0 cm. What is the z-score of a woman who is 159.0 cm tall?
z = -0.3968
(b) The manufacturer designs the seats to fit women with a maximum
z-score of 2.8. How tall is a woman with a z-score of 2.8? 179.14 = x

5. Curving The Marks
Professor Adams has 140 students who wrote a statistics test. If the
marks are approximately normally distributed:
• how many students should have a 'B' mark (i.e., 70 to 79 percent)?
• how many students should have failed (i.e., less than 50 percent)?
• how high must she set the mark for an 'A' if she wants 5
percent of the students to get an A?
• how high must she set the passing mark if she wants only the
top 75 percent of the marks to be passing marks?

6. Working with the Normal Distribution
Example: The graph at right
represents the marks of a large
number of students where the mean
mark is 69.3 percent and the
standard deviation is 7 percent, and
the distribution of marks is
approximately normal.
62.3% 69.3% 76.3%

7. Working with the Normal Distribution
Example: The graph at right represents
the marks of a large number of students
where the mean mark is 69.3 percent
and the standard deviation is 7 percent,
and the distribution of marks is
approximately normal.
We know that if the marks are approximately normally distributed, then
approximately:
• 68 percent of the marks are between 62.3 percent and 76.3 percent (i.e., µ ± 1σ)
• 34% of the marks are between 69.3 percent and 76.3 percent (i.e., between µ
and µ + 1σ)
• 50% of the marks are below 69.3 percent (i.e., less than µ)
• 16% of the marks are above 76.3 percent (i.e., greater than µ + 1σ)

8. The Standard Normal Distribution
The normal distributions used up to this point have all been non-
standard normal distributions. This means that the mean and
standard deviation of the distribution are the mean and standard
deviation of the data being studied.
In a standard normal distribution, the scale on the x-axis is the z-
score (a standard score) where 'z = 0' is the mean, and the
standard deviation is '1'. The distribution is a probability
distribution where the area under the curve = 1. This means
there is a 100 percent chance of every score being included in
this distribution. Other probabilities are illustrated below.

9. ShadeNorm(Lo z, Hi z)
[shades area under std. normal curve]
σ
ShadeNorm(Lo value, Hi value, mean, std. dev.)
[shades area under modified normal curve]
µ-5σ
σ
µ+5σ

10. Case 1(a): Calculate the Percentage of Scores Between Two Given
Scores
HOMEWORK
The mean mark for a large number of students is 69.3 percent
with a standard deviation of 7 percent. What percent of the
students have a 'B' mark (i.e., 70 percent to 79 percent)?
Assume that the marks are normally distributed.

11. HOMEWORK
Find the percent of z-scores in a standard normal distribution that are:
(a) below z = 0.52 (b) above z = -2.35
(d) between z = 0.55 and z = 0.15
(c) between z = -1.11 and z = 0.92

12. HOMEWORK
Find the z-score if the area under a standard normal curve:
(a) to the left of z is 0.812
(b) to the right of z is 0.305
(c)to the right of z is 0.785

13. HOMEWORK
Determine the values for z and x.