Transcript Statistical Analysis – Chapter 4 Normal Distribution

Statistical Analysis – Chapter 4
Normal Distribution
What is the normal curve?

In chapter 2 we talked about histograms and modes

A normal distribution is when a set of values for one variable,
when displayed in a histogram (or line graph) has one peak
(mode) and looks like a bell. Here is an example using height:
Characteristics of the Normal Curve
a.
Bell shaped, fading at the tails. In other words, more values
are in the middle, and odd or unusual values fall at the tails
b.
All (100%) of the data fits on the curve, with 50% before the
mean and 50% after
c.
68% of the data falls within -1 and +1 standard deviations of
the mean
d.
95% of the data falls between -2 and +2 standard deviations
e.
The percentage of data between any two points is equal to
the probability of randomly selecting a value between the
two points (remember classical probability from Ch. 3)
Standard Deviations and Z-Score

Z – scores = the number of standard deviations away from the
mean.

z-score = x - µ
σ
(x = data for which we want to know the z-score)

We use the characteristics of the normal curve, and the zscore, to find out the probability of a particular event or value
occurring (remember classical probability from Chapter 3)
Solving Normal Curve Problems Using
Z-Scores
(steps listed at bottom of p. 111)
1.
2.
3.
4.
Draw a normal curve, showing values for (-2 through
+2)
Shade the area in question
Calculate the z scores and cutoffs (percentages asked
for)
Use the z-scores and cutoffs to solve the normal curve
problem
Find Percentages on the Normal Curve
Table
Let’s do these questions as a class…
a.
b.
c.
d.
e.
What is the percentage of data from z = 0 to z = 0.1?
What is the percentage of data from z = 0 to z = 2.16?
What is the percentage of data from z = -1.11 to z = 1.11?
What is the percentage of data above z = 1.24?
What is the percentage of data below z = -0.6?
Answers
a.
b.
c.
d.
e.
.0398…39.8%
.4846…48.46%
.3665 + .3665 = .733…73.3%
.50 - .3925 = .1075…10.75%
.50 - .2257 = .2743…27.43%
Working backwards from percentages…

When working backwards from percentages, we still use
the normal table…but look for the percentage to give us
the z-score…
a.
What is the z-score associated 10.2% of the data?
What is the z-score(s) for the middle 30% of the normal
curve?
What is the z-score of data in the upper 25% of the normal
curve?
b.
c.
Answers
a.
b.
c.
z = 0.26
z = -.39 to z = .39
z = 0.67
Let’s do Question 4.2
Use the normal curve table to determine the percentage of data in the
normal curve
a.
Between z = 0 and z = .82
b.
Above z = 1.15
c.
Between z = -1.09 and z = .47
d.
Between z = 1.53 and z = 2.78
Work backward in the normal curve table to solve the following:
e.
f.
g.
32% of the data in the normal curve data can be found between z = 0 and
z=?
Find the z score associated with the lower 5% of the data.
Find the z scores associated with the middle 98% of the data.
Question 4.2 Answers
Answers to Question 4.2
a. 29.39%
b. 12.51%
c. 54.29%
d. 6.03%
e. Between z = 0 and z = .92, or between z = 0 and z =
-.92
Question 4.7
Use the normal curve table to determine the percentage of data in the
normal curve
a.
Between z = 0 and z = .38
b.
Above z = -1.45
c.
Above z = 1.45
d.
Between z = .77 and z = 1.92
e.
Between z = -.25 and z = 2.27
f.
Between z = -1.63 and z = -2.89
Work backward in the normal curve table to solve the following.
g.
15% of the data in the normal curve can be found between z = 0 and z =
?
h.
Find the z score associated with the upper 73.57% of the data.
i.
Find the z scores associated with the middle 95%
Question 4.7 Answers
a.
b.
c.
d.
e.
f.
g.
h.
i.
14.80%
92.65%
7.35%
19.32%
58.71%
4.97%
z = .39 or -.39
z = -.63
Between z = -1.96 and z = +1.96
Binomial Distributions and Sampling
Binomial means two categories in a population…
 Males and females
 Sports game players vs. Non sports game players
 Incomes over 40,000 vs. incomes under 40,000
Quick note: Remember…for binomial distributions, we would
visualize this data through a pie chart…because we do not
have enough categories for a histogram…
Sampling from a Two-Category
Population

With two-category populations, we can describe the
population by p – the percentage of values in one
category

This is the same p from the last chapter on probability
(classical probability)…
P(event) ≈
s (number of chances for success)
n (total equally likely possibilities)

We know (actually….statisticians know) that if we
randomly sampled from a population, then
ps ≈ p
Sampling Distribution

In order to know the odds of getting certain values from
this particular binomial sample, we have to know the
sampling distribution from this population.

Under certain conditions, the sampling distribution for a
binomial value is normal (i.e. the distribution follows the
normal curve).

When the sampling distribution is normal, then we can
make predictions using our table and our z-scores
Sampling from a Binomial Distribution

Suppose, we defined a population (full time FIT students
who either shop at Hot Topic), and we have made our
measure of interest into a binomial distribution – those
who shop at Hot Topic and those who do not.

Suppose over the last 10 years, marketers have surveyed
the FIT population hundreds of times and found that Hot
Topic shoppers are p = .13. (those who are non-Hot Topic
shoppers is p = .87)
Sampling from a Binomial Distribution

But suppose sometime later, your manager asks you
to lead another study. But this time, you don’t have
enough money to survey the whole population, and
you have to get a sample.

We can assume, because so many studies have been
done in the past that the true value of Hot Topic
shoppers is p = .13. Thus, because we know that ps ≈
p, your sample should have approximately the same
value.
Sampling from a Binomial Distribution

For each sample, we can use the number sampled, and the p
value from the population to predict the total number of Hot
Topic shoppers. This is called the expected value.

Expected value = np

Thus, if we collected a sample of 200 FIT students, how many
students would we expect to be Hot Topic shoppers?
np = (200)(.13) = 26

This expected value is the mean of your sample
Binomial Distribution and the Normal
Curve

Now, we need to decide if we can use the normal curve to
solve problems…

If (np) > 5 and n(1 – p)>5…then the sampling distribution will
be normally distributed.

So, our sample was 200 students.
Is (np) > 5?
Is n(1 – p)>5?
Yes…and yes.
np = (200)(.13) = 26
n(1 – p) = (200)(1 - .13) = (200)(.87) = 174

Binomial Distribution and the Normal
Curve

What do we mean that a sampling distribution is normal?
a)

Just like someone’s age is one value among many ages that we tally
to make a histogram, we can tally many samples, get the p values of
those sample, and construct histograms from these means.
If we took say, 1000 samples, and tallied the p values for Hot
Topic shoppers, then those values, when turned into a
histogram, should form a normal curve. Just like if we took
the heights of a 1000 women, and tallied those values to get a
normal curve.
How to use the Binomial Distribution
and the Normal Curve
1.
2.
3.
4.
5.
6)
Get the mean (µ)…the mean is the expected value
(np)
Get the standard deviation (σ) = √np(1 – p)
Draw a normal curve using mean and standard dev
Use the “continuity correction factor,” and add +/half a unit to the value we want to solve for
Get the z-scores = x - µ
σ
Use the normal curve table to solve the problem
Why the “continuity correction factor”?

This is only for discrete values (where values occupy only distinct points.) For
example, in our study, there is no such thing as a “half” or “3/4” Hot Topic shopper.
Either you are a shopper or not. Looking at how histograms are presented, you
can see why we have to use the correction factor.
1. Probability of getting a value equal
to or greater than (=>), then you
must subtract a half-unit
2. Probability of getting a value equal
to or lesser than (=<), you must
add a half unit.
3. Probability of getting the exact
value, you must get the Z-scores
for a half-unit above and a halfunit below
Now let’s answer a Hot Topic
Question…
If you collected a sample of 200 FIT students…
a. What is the probability that 13 will be Hot
Topic shoppers?
b. What is the probability that you will have 30
or more Hot Topic shoppers?
c. What is the probability that you will have 25
or less Hot Topic shoppers?
Question
1.
What is the probability that 13 will be Hot Topic shoppers?
2.
What is the probability that you will have 30 or more Hot Topic
shoppers?
3.
What is the probability that you will have 25 or less Hot Topic shoppers?
Answer
1.
Get the mean (µ) = expected value = np = (200)(.13) = 26
2.
Get the standard deviation (σ) = √np(1 – p) = √26(1 - .13) = √26(.87) =
√22.62 ≈ 4.76
3.
Draw a normal curve using mean and standard dev.
4.
Use the continuity correction factor to correct x. (a) 12.5 and 13.5, (b)
29.5, (c) 25.5
5.
Get the z-scores. (a) -2.83 and -2.62, (b) .735, (c)-.105
6.
Solve the problem… (a) 4977 - .4956 = .002, or 2% (b) .50 - .2704 ≈ .23,
or 23%, (c) .50 - .0596 = .4404
Now let’s do question 4.16 as a class…
In a marketing population of phone calls, 3% produced a sale. If
this population proportion (p = 3%) can be applied to future
phone calls, then out of 500 randomly monitored phone calls,
a.
b.
c.
How many would you expect to produce a sale?
What is the probability of getting 11 to 14 sales?
What is the probability of getting 12 or less sales?
a. 15
b. 32.93%
c. 25.46%
Question 4.16 answers
a.
b.
c.
Expected value = np = 500(.03) = 15
32.93%
25.46%