Finite Mathematics Lesson 10

 Section 7.5 Section 7.6

Chapters 7.5 - 7.6

The variance and the standard deviation are measures that describe the type of data that was collected. Most scientific and business calculators can calculate these statistics.   Spreadsheets such as Excel have extensive statistical features you may want to explore.  I  made some interactive tutorials for these sections.  Check them out on the Course Content page.

Course Notes

Section 7.5   The Variance and Standard Deviation

The variance is a measure of how spread out data is about its mean.  If most of the data points are close to its mean, the smaller the variance.  If most of the data points are spread out relative to its mean, the larger the variance.  Lets compare two different populations that have the same mean.  These are the results from the first two test given in a math course.

 Test 1 Test 2 20 50 40 55 60 60 80 65 100 70
 µTest1 = (20 + 40 +60+ 80 + 100) ÷ 5 = 60 µTest2 = (50 + 55 +60+ 65 + 70) ÷ 5 = 60

As you can see form the test scores above, even though the means (averages) are identical, they do not reflect the same experience. So lets look at another parameter of the population called the variance.   (denoted by the lowercase Greek letter sigma squared, s² )

• The variance for Test 1
• The variance for Test 2

The variances for each population are quite different. Remember, the smaller variance indicates closer the values are to the mean. The standard deviation is used more often than the variance to interpret data.  The symbol for the standard deviation is the Greek letter sigma, s.  It is smaller than the variance and is easily interpreted.  (Chapter 7.5)

• The standard deviation for Test 1
• The standard deviation for Test 2

The variance of the probability distribution is calculated differently.  Instead of dividing  the sum of the terms ,(xi - µ)2 , by N,  we multiply each term, (xi - µ)2 ,  by the its corresponding probability.
[variance] = (x1 - µ)2 p1 + (x2 - µ)2 p2 + ··· + (xN - µ)2 pN

Example  Find the variance and standard deviation of ...

 Outcome Probability -2 0.3 0 0.4 2 0.2 12 0.1
• First we need to find the mean.
µ = (-2)(0.3) + 0(0.4) + 2(0.2) + 12(0.1) = -0.6 + 0 + 0.4 + 1.2 = 1
• Next we need to find the variance. (x1 - µ)2 p1 + (x2 - µ)2 p2 + ··· + (xN - µ)2 pN (-2 - 1)2 0.3 + (0 - 1)2 0.4 +  (2 - 1)2 0.2 +  (12 - 1)2 0.1 9(0.3) + 1(0.4) + 1(0.2) + 121(0.1) = 15.4
• Now we can find the standard deviation.

Important Remark
The formulas for the variance of a population and the variance of a sample are calculated differently. Recall the sample is not the entire population.

 sample variance population variance

So the standard deviation are different also...

 sample standard deviation population standard deviation

Chebychev's Inequality
Chebychev's Inequality helps us determine the likelihood of having extreme values in the data.  Suppose a probability distribution has mean 40 and standard deviation 2.

 Test 1 Chebychev's Inequality µ =  40 The probability that a randomly chosen outcomes lies between  µ - c  and   µ + c  is at least

Estimate the probability that the outcome is  between 30 and 50 .

• First, find c given µ -and   µ + c .
40 - c = 30  and  40 + c = 50  gives  c = 10
• Therefore, we expect at least 96% of the outcomes to be between  30 and 50.

Section 7.6   The Normal Distribution

The most important distribution in statistics, a normal (or Gaussian) distribution, has probabilities that follow the familiar bell-shaped curve:

A normal distribution is completely specified by giving:

• its mean value, frequently denoted by m,
• its standard deviation, frequently denoted by s.
 Large s Small s

Suppose a random variable has a normal distribution:

 A value of that quantity is just as likely to lie above the mean as below it (this is why the bell curve is centered on m). A value of that quantity is less likely to occur the farther it is from the mean (this is why the bell curve decreases in both directions from m). Values to one side of the mean are of the same probability as values at the same distance on the other side of the mean (this is why the bell curve is symmetric about m). Values at a distance greater than 3s from m are possible but very unlikely (this is why the bell curve appears to hit the horizontal axis).

How do we use the standard deviation in the normal curve?
The total area under the curve is 1.

• The probability that the outcome is within 1 standard deviation of the mean is written as:
Pr ( µ - 1s < X < m + 1s) = 0.683
• The probability that the outcome is within 2 standard deviation of the mean is written as:
Pr ( µ - 2s < X < m + 2s) = 0.954
• The probability that the outcome is within 3 standard deviation of the mean is written as:
Pr ( µ - 3s < X < m + 3s) = 0.997

The total area under the curve is 1.  So for a normal distribution curve, over 68% of the area is contained by the values that are 1 standard deviation on either side of the mean, over 95% with 2 standard deviations, and over 99% with 3 standard deviations!

Let's look at our first example again.

 From  Test 2 above   µ = 60 and s =7 We will find: Pr ( µ - 1s < X < m + 1s) = Pr ( µ - 2s < X < m + 2s) = Pr ( µ - 3s < X < m + 3s) = Pr ( µ - 1s < X < m + 1s) = µ  + 1s = 60 + 1(7) = 67 µ  - 1s = 60 - 1(7) = 53 Pr ( 53< X < 67) = 0.683 Note k =1 Pr ( µ - 2s < X < m + 2s) = µ  + 2s = 60 + 2(7) = 74 µ  - 2s = 60 - 2(7) = 46 Pr ( 46< X < 74) = 0.954 Note k =2 Pr ( µ - 3s < X < m + 3s) = µ  +3s = 60 +3(7) = 39 µ  - 3s = 60 - 3(7) = 81 Pr ( 39< X <  81) = 0.997 Note k =3

Remark
When talking about testing, saying a person is within 1 standard deviation of the mean implies that the person is with 68% of the of their classmates.  If you were 3 standard deviations above the mean,  you world be in the top 2.5% of the students.   If you were 3 standard deviations below the mean,  you world be in the bottom 2.5% of the students.

Using the Normal distribution
If you hadn't noticed in the book,  the formula for calculating the area under the normal curve is gnarly. Look at Table 1 in Appendix A in the back of your book.  There is an abbreviated version of it on page 372.  Since the properties of the normal curve are the same regardless of its mean and standard deviation,  we can use one table if we let µ = 0 and s = 1.

A(1.5) - A(-.5) = 0.9332 - 0.3085 = 0.6247

Find the area up to 1.5, then subtract the area below -0.5 to get the area of the shaded region.  Look up A(1.5) and A(-.5) in appendix A.

Converting to the Normal Curve

Most of the time the standard deviation s is not 1 or the mean m is not 0.  Here is an easy way to convert to the Normal curve.

or written as    z = (x - m) ÷ s

Example #28, page 380
Suppose that IQ scores are normally distributed with m = 100 and s = 10.  What percent of the population have IQ scores greater than 125 or more?

• z = (x - m) ÷ s = (125 - 100) ÷ 10 = 25  ÷ 10 = 2.5
• So  1 - A(z) = 1  - A(2.5) = 1  - 0.9938 = 0.0062 = 0.62%

According to this problem,  less than 1 % of the population has an IQ of 125 or over.

µ = 60 and s =7  and z = (x - m) ÷ s
z = (67 - 60) ÷ 7 = 1
z = (53 - 60) ÷ 7 = -1

Pr ( -1< Z < 1) = A(1) - A(-1) = .8413 - .1587 = .6826
We arrive at the same value!

Remark
You don't have to remember the formulas on page 352 if you can remember how to convert to a normal curve.
Also I used < symbol instead of less than or equal signs to cut down on download time.   See note on page 348 at bottom of page.

or written as    z = (x - m) ÷ s

Section 7.7   Normal Approximation to the Binomial Distribution

Read this section. This is actually useful.  Look at example #2 on page 383 for a relevant math problem.  Here the mean m = np and   standard deviation is s = squareroot(npq).  So if you know the distribution is binomial, you can use the normal curve  to approximate binomial probability distributions. Just do #1 and #3 on page 385 for practice.