Transcript No Slide Title

Chapter 6 The Normal
Probability Distribution
General Objectives:
You will learn about continuous random variables and their
probability distributions. You will learn how to calculate normal
probabilities and, under certain conditions, how to use the
normal probability distribution to approximate the binomial
probability distribution.
©1998 Brooks/Cole Publishing/ITP
Specific Topics
1. Probability distributions for continuous random variables
2. The normal probability distribution.
3. Calculation of areas associated with the normal probability
distribution
4. The normal approximation
©1998 Brooks/Cole Publishing/ITP
6.1 Probability Distributions for
Continuous Random Variables

Continuous random variables can assume the infinitely many
values corresponding to points on a line interval.
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Examples: Heights, weights, length of life of a particular product,
experimental laboratory error

A smooth curve describes the probability distribution of a
continuous random variable.

The depth or density of the probability, which varies with x,
may be described by a mathematical formula f (x ), called the
distribution for the random variable x.
©1998 Brooks/Cole Publishing/ITP

Several important properties of continuous probability
distributions:
- The area under a continuous probability distribution is equal
to 1.
- The probability that x will fall into a particular interval, say,
from a to b, is equal to the area under the curve between the
two points a and b.
- P (x = a) = 0 for continuous random variables.
- This implies the P (x  a) = P (x > a) and P (x  a) = P (x < a).
- This is not true in general for discrete random variables.

Figure 6.1 shows relative frequency histograms for increasingly
large sample sizes. Figure 6.2 shows a probability distribution
f (x ) and the region P (a < x < b ).
©1998 Brooks/Cole Publishing/ITP


Choosing the probability distribution f (x ) appropriate for a given
experiment:
- It fits the accumulated body of data.
- It allows us to make the best possible inferences using
the data.
The normal probability distribution provides a good model for
describing data that have mound-shaped frequency
distributions.
The Normal Probability Distribution:
2
1



x

m
/ 2s 2 
f (x) =
e
s 2p

where e = 2.718 and p = 3.142; m and s (s > 0 ) are the
parameters that represent the population mean and standard
deviation.
Figure 6.3 shows the normal probability distribution.
©1998 Brooks/Cole Publishing/ITP

Figure 6.4 shows several such distributions with differing values
of m and s .
Figure 6.4



m locates the center of the distribution and the distribution is
symmetric about m .
The areas to the right and left of m are both .5.
The shape of thesdistribution is determined by s, the population
standard deviation.
©1998 Brooks/Cole Publishing/ITP



Large values of s reduce the height of the curve and increase
the spread.
Small values of s increase the height of the curve and reduce
the spread.
Many positive random variables, such as height, weight, and
time, have distributions that are well approximated by a normal
distribution.
©1998 Brooks/Cole Publishing/ITP
6.3 Tabulated Areas of the
Normal Probability Distributions


The probability that a continuous random variable x assumes a
value in the interval a to b is the area under the probability
density function between the points a and b.
We use a standardization procedure that allows us to use the
same tables of probabilities for all normal distributions.
The Standard Normal Random Variable:
A normal random variable is standardized by expressing its
value as the number of standard deviations (s ) it lies to the left
or right of its mean m. The standardized normal random variable
z, is defined as z = (x  m)/ s , or equivalently, x = m + zs .
©1998 Brooks/Cole Publishing/ITP

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From the formula for z, we can draw these conclusions:
- When x is less that the mean m , the value of z is negative.
- When x is greater that the mean m , the value of z is positive.
- When x = m , the value z = 0.
The standard probability distribution has a mean of zero and a
standard deviation of 1.
The area under the standard normal curve between mean z = 0
and a specified positive value of z, say, z0 , is the probability
P (0  z  z0).
This area is recorded in a table where z, correct to the nearest
tenth, is recorded in the left-hand column of the table, and the
second decimal place for z, corresponding to hundredths, is
given across the top row.
©1998 Brooks/Cole Publishing/ITP

Figure 6.5 shows the standardized normal distribution.
Table 6.1 gives an abbreviated version of Table 3 in Appendix I.

Examples 6.1–6.4 illustrate the use of the standard normal
curve.
Figure 6.5
©1998 Brooks/Cole Publishing/ITP
Table 6.1 Abbreviated version of Table 3 in Appendix I
©1998 Brooks/Cole Publishing/ITP
Example 6.1
Find P (0  z  1.63). This probability corresponds to the area
between the mean (z = 0) and a point z = 1.63 standard
deviations to the right of the mean (see Figure 6.6).
Solution
The area is shaded in Figure 6.6. Since Table 3 in Appendix I
gives areas under the normal curve to the right of the mean, you
need only find the tabulated value corresponding to z = 1.63.
Proceed down the left-hand column of the table to z = 1.6 and
across the top of the table to the column marked .03. The
intersection of this row and column combination gives the area
A = .4484. Therefore, P (0  z  1.63) = .4484.
©1998 Brooks/Cole Publishing/ITP
Example 6.2
Find P (.5  z  1.0). This probability corresponds to the area
between z = .5 and z = 1.0, as shown in Figure 6.7.
Solution:
The area required is equal to the sum of A 1 and A 2 , shown in
Figure 6.7. From Table 3 in Appendix I, you find A 2 = .3413.
The area A 1 equals the area between z = 0 and z = .5, or
A 1 = .1915. Thus, the total area is
A = A 1 + A 2 = .1915 + .3413 = .5328
That is, P (.5  z  1.0) = .5328.
©1998 Brooks/Cole Publishing/ITP

How can you find areas to the left of the mean? Since the
standard normal curve is symmetric about z = 0, any area to the
left of the mean is equal to the equivalent to the right of the
mean.
Calculating Probabilities for a
General Normal Random Variable:
- Most of the time a normal random variable is involved.
- Standardize the interval of interest, writing it as the equivalent
interval in terms of z.
- The probability of interest is the area that you find using the
standard normal probability distribution.

Examples 6.5 – 6.7 illustrate the conversion of various normal
distributions to the standardized distributions.
©1998 Brooks/Cole Publishing/ITP
6.4 The Normal Approximation
to the Binomial Probability
Distribution (Optional)

Two ways to calculate probabilities for the binomial random
variable x:
- Using the binomial formula,
P ( x = k ) = Ckn p k q n k
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
- Using the cumulative binomial tables
There is one other option when np < 7; the Poisson probabilities
can be used to approximate P(x = k).
When this does not work and n is large, the normal probability
distribution provides another approximation for binomial
probabilities.
©1998 Brooks/Cole Publishing/ITP
The Normal Approximation to the
Binomial Probability Distribution:
- Let x be a binomial random variable with n trials and probability
p of success.
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- The probability distribution of x is approximated using a normal
curve with m = np and s = npq.
- This approximation is adequate as long as n is large and p is
not too close to 0 or 1.
You must be careful not to exclude half of the two extreme
probability rectangles when you use the approximation.
This adjustment, called the continuity correction, helps account
for the fact that you are approximating a discrete random
variable with a continuous one.
Figure 6.12 shows the binomial probability distribution for n = 25
and p = .5 and the approximating normal distribution with
m = 12.5 and s = 2.5. Figure 6.13 shows the binomial probability
distribution for n = 25 and p = .1.
©1998 Brooks/Cole Publishing/ITP

Example 6.8 uses the normal curve to approximate binomial
probabilities.
Example 6.8
Use the normal curve to approximate the probability that x = 8,
9, or 10 for a binomial random variable with n = 25 and p = .5.
Compare this approximation to the exact binomial probability.
Solution
You can find the exact binomial probability for this example
because there are cumulative binomial tables for n = 25. From
Table 1 in Appendix I,
P ( x = 8, 9, or 10) = P ( x  10)  P ( x  7) = .212  .022 = .190
To use the normal approximation, first find the appropriate
mean and standard deviation for the normal curve:
m = np = 25(.5) = 12.5
s = npq = 25(.5)(.5) = 6.25 = 2.5
©1998 Brooks/Cole Publishing/ITP
The probability that you need corresponds to the area of the
three rectangles lying over x = 8, 9, and 10. The equivalent area
under the normal curve lies between x = 7.5 (the lower edge of
the rectangle for x = 8 ) and x = 10.5 (the upper edge of the
rectangle for x = 10 ). This area is shaded in Figure 6.12.
To find the normal probability, follow the procedures of
Section 6.3. First you standardize each interval endpoint:
z=
xm
s
xm
=
7.5  12.5
= 2.0
2 .5
10.5  12.5
= .8
s
2 .5
The area is shaded in Figure 6.14. Then the approximate
probability is found from Table 3 in Appendix I:
z=
=
P ( 2.0 < z < .8 ) = .4772  .2881 = .1891
©1998 Brooks/Cole Publishing/ITP
Figure 6.12 The binomial probability distribution for n = 25 and
p = .5 and the approximating normal distribution with m = 12.5
and s = 2.5
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Figure 6.14 Area under the standard normal curve
for Example 6.8
©1998 Brooks/Cole Publishing/ITP

Be careful not to exclude half of the two extreme probability
rectangles when you use the normal approximation to the
binomial probability distribution. This is called the continuity
correction.
Rule of Thumb:
The normal approximation to the binomial probabilities will be
adequate if both np > 5 and nq > 5.
©1998 Brooks/Cole Publishing/ITP
Calculating Binomial Probabilities Using the
Normal Approximation:
- Find the necessary values on n and p.
Calculate m = np and s = npq .
- Write the probability you need in terms of x and locate the
appropriate area of the curve.
- Correct the value of x by .5 to include the entire block of
probability for that value. This is the continuity correction.
- Convert the necessary x values to z values using
x + .5  np
npq
- Use Table 3 in Appendix I to calculate the approximate
probability.
z=

Example 6.9 applies the normal approximation to the binomial,
as does Example 6.10.
©1998 Brooks/Cole Publishing/ITP-
Example 6.9
The reliability of an electrical fuse is the probability that a fuse,
chosen at random from production, will function under the
conditions for which it has been designed. A random sample of
1000 fuses was tested and x = 27 defectives were observed.
Calculate the approximate probability of observing 27 or more
defectives, assuming that the fuse reliability is .98.
Solution
The probability of observing a defective fuse is tested is p = .02,
given that the fuse reliability is .98. Then
m = np = 1000(.02) = 20
s = npg = 1000(.02)(.98) = 4.43
The probability of 27 or more defective fuses, given n = 1000, is
P ( x  27) = p(27) + p(28) + p(29) +  + p(999) + p(1000)
©1998 Brooks/Cole Publishing/ITP
It is appropriate to use the normal approximation to the binomial
probability because
np = 1000(.02) = 20 and nq = 1000(.98) = 980
are both greater than 5. The normal area used to approximate
P (x  27) is the are under the normal curve to the right of 26.5,
so that the entire rectangle for x = 27 is included. Then, the
z-value corresponding to x = 26.5 is
z=
xm
s
=
26.5  20 6.5
=
= 1.47
4.43
4.43
and the area between z = 0 and z = 21.47 is equal to .4292, as
shown in Figure 6.15. Since the total area to the right of the
mean is equal to .5, you have
P (x  27)  P (z  1.47) = .5  .4292 = .0708
©1998 Brooks/Cole Publishing/ITP
Key Concepts and Formulas
I. Continuous Probability Distributions
1. Continuous random variables
2. Probability distributions or probability density functions
a. Curves are smooth.
b. The area under the curve between a and b represents
the probability that x falls between a and b.
C. P (X = a) = 0 for continuous random variables.
II. The Normal Probability Distribution
1. Symmetric about its mean m .
2. Shape determined by its standard deviation s .
©1998 Brooks/Cole Publishing/ITP
III. The Standard Normal Distribution
1. The normal random variable z has mean 0 and standard
deviation 1.
2. Any normal random variable x can be transformed to a
standard normal random variable using
xm
z=
s
3. Convert necessary values of x to z.
4. Use Table 3 in Appendix I to compute standard normal
probabilities.
5. Several important z-values have tail areas as follows:
Tail Area:
z-Value:
.005
2.58
.01
2.33
.025
1.96
.05
1.645
.10
1.28
©1998 Brooks/Cole Publishing/ITP