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## Chapter 6 ~ Normal Probability Distributions

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Title: Chapter 6 ~ Normal Probability Distributions

1
Chapter 6 Normal Probability Distributions
2
Chapter Goals
• Learn about the normal, bell-shaped, or Gaussian
distribution
• How probabilities are found
• How probabilities are represented
• How normal distributions are used in the real
world

3
6.1 Normal Probability Distributions
• The normal probability distribution is the most
important distribution in all of statistics
• Many continuous random variables have normal or
approximately normal distributions
• Need to learn how to describe a normal
probability distribution

4
Normal Probability Distribution
• 1. A continuous random variable

2. Description involves two functions a. A
function to determine the ordinates of the graph
picturing the distribution b. A function to
determine probabilities
4. The probability that x lies in some interval
is the area under the curve
5
The Normal Probability Distribution
6
Probabilities for a Normal Distribution
7
Notes
• The definite integral is a calculus topic
• We will use the TI83/84 to find probabilities for
normal distributions
• We will learn how to compute probabilities for
one special normal distribution the standard
normal distribution
• We will learn to transform all other normal
probability questions to this special distribution
• Recall the empirical rule the percentages that
lie within certain intervals about the mean come
from the normal probability distribution
• We need to refine the empirical rule to be able
to find the percentage that lies between any two
numbers

8
Percentage, Proportion Probability
• Basically the same concepts
• Percentage (30) is usually used when talking
about a proportion (3/10) of a population
• Probability is usually used when talking about
the chance that the next individual item will
possess a certain property
• Area is the graphic representation of all three
when we draw a picture to illustrate the situation

9
6.2 The Standard Normal Distribution
• There are infinitely many normal probability
distributions
• They are all related to the standard normal
distribution
• The standard normal distribution is thenormal
distribution of the standard variable z(the
z-score)

10
Standard Normal Distribution
• Properties
• The total area under the normal curve is equal to
1
• The distribution is mounded and symmetric it
extends indefinitely in both directions,
approaching but never touching the horizontal
axis
• The distribution has a mean of 0 and a standard
deviation of 1
• The mean divides the area in half, 0.50 on each
side
• Nearly all the area is between z -3.00 and z
3.00
• Notes
• Table 3, Appendix B lists the probabilities
associated with the intervals from the mean (0)
to a specific value of z
• Probabilities of other intervals are found using
the table entries, addition, subtraction, and
the properties above

11
Table 3, Appendix B Entries
• The table contains the area under the standard
normal curve between 0 and a specific value of z

12
Example
• Example Find the area under the standard normal
curve between z 0 and z 1.45

13
Using the TI 83/84
• To find the area between 0 and 1.45, do the
following
• 2nd DISTR 2 which is normalcdf(
• Enter the lower bound of 0
• Enter a comma
• Then enter 1.45
• Close the parentheses if you like or hit Enter
• The value of .426 is shown as the answer!
• Interpretation of the result The probability
that Z lies between 0 and 1.45 is 0.426

14
Example
• Example Find the area under the normal curve to
the right of z 1.45 P(z gt 1.45)

15
Using the TI 83/84
• To find the area between 1.45 and 8, do the
following
• 2nd DISTR 2 which is normalcdf(
• Enter the lower bound of 1.45
• Enter a comma
• Then enter 1 2nd EE 99
• Close the parentheses if you like or hit Enter
• The value of .074 is shown as the answer!
• Interpretation of result The probability that Z
is greater than 1.45 is 0.074

16
Example
• Example Find the area to the left of z 1.45
P(z lt 1.45)

17
Using The TI 83/84
• To find the area between - 8 and 1.45, do the
following
• 2nd DISTR 2 which is normalcdf(
• Enter the lower bound of -1 2nd EE 99
• Enter a comma
• Then enter 1.45
• Close the parentheses if you like or hit Enter
• The value of 0.926 is shown as the answer!
• Interpretation of result The probability that Z
is less than 1.45 is 0.926

18
Notes
• The addition and subtraction used in the previous
examples are correct because the areas
represent mutually exclusive events
• The symmetry of the normal distribution is a key
factor in determining probabilities associated
with values below (to the left of) the mean. For
example the area between the mean and z -1.37
is exactly the same as the area between the mean
and z 1.37.
• When finding normal distribution probabilities, a

19
Example
• Example Find the area between the mean (z 0)
and z -1.26

20
Using the TI 83/84
• Find the area to the left of z -0.98
• Use -1E99 for - 8 and enter 2nd DISTR
• Normalcdf (-1e99, -0.98) which gives .164

21
Example
• Example Find the area between z -2.30 and z
1.80

22
Using the TI 83/84
• Find the area between z -2.30 and z 1.80
• Enter 2nd DISTR, normalcdf (-2.3, 1.80) and press
enter
• .953 is given as the answer.
• Remember, the function normalcdf is of the form
• Normalcdf(lower limit, upper limit, mean,
standard deviation) and if youre working with
distributions other than the standard normal
(recall mean 0, stddev 1), you must enter the
values for mean and standard deviation

23
Normal Distribution Note
• The normal distribution table may also be used to
determine a z-score if we are given the area
(working backwards)
• Example What is the z-score associated with the
85th percentile?

24
Using the TI 83/84
• There is another function in the DISTR list that
is used to find the value of z (or x) when the
probability is given. For the previous problem,
we are actually asking what is the value of z
such that 85 of the distribution lies below it.

25
Using the TI 83/84
• Use 2nd DISTR invNorm( to calculate this value
• 2nd DISTR invNorm(.85) ENTER gives us a value
of 1.036 which is shown

26
Example
• Example What z-scores bound the middle 90 of a
standard normal distribution?

27
Using the TI 83/84
• The TI 83/84 calculates areas from -8 to the
value of z we are interested in. Therefore, we
must get a little creative to solve some
problems.
• Using the idea that the total area equals one
comes in very handy here!
• For the example given, where we are interested in
the value of z that bounds the middle 90, the
tails therefore represent a total of 10. Divide
this in two since it is symmetric and this gives
5 in each tail.

28
Using the TI 83/84
• Now use the 2nd DISTR invNorm with .05 in the
argument like this
• Which gives an answer of -1.645
• Since the distribution is symmetric, the upper
limit is 1.645, so 90 of the distribution lies
between
• (-1.645, 1.645)

29
Using the TI 83/84
• Now lets work the problems on page 279

30
6.3 Applications of Normal Distributions
• Apply the techniques learned for the z
distribution to all normal distributions
ofx-values
• Convert, or transform, the question into an
equivalent probability statement
involvingz-values

31
Standardization
• Suppose x is a normal random variable with mean m
and standard deviation s

32
Example
• Example A bottling machine is adjusted to fill
bottles with a mean of 32.0 oz of soda and
standard deviation of 0.02. Assume the amount
of fill is normally distributed and a bottle is
selected at random

1) Find the probability the bottle contains
between 32.00 oz and 32.025 oz 2) Find the
probability the bottle contains more than 31.97 oz
33
Solution Continued
34
Example, Part 2
2)
35
Notes
• The normal table may be used to answer many kinds
of questions involving a normal distribution
• Often we need to find a cutoff point a value of
x such that there is a certain probability in a
specified interval defined by x
• Example The waiting time x at a certain bank is
approximately normally distributed with a mean
of 3.7 minutes and a standard deviation of 1.4
minutes. The bank would like to claim that 95
of all customers are waited on by a teller
within c minutes. Find the value of c that
makes this statement true.

36
Solution
37
Example
• Example A radar unit is used to measure the
speed of automobiles on an expressway during
rush-hour traffic. The speeds of individual
automobiles are normally distributed with a mean
of 62 mph. Find the standard deviation of all
speeds if 3 of the automobiles travel faster
than 72 mph.

38
Solution
-
m
x

z

s
.

1
88
10
s
39
Notation
• If x is a normal random variable with mean m and
standard deviation s, this is often denoted x
N(m, s)
• Example Suppose x is a normal random variable
with m 35 and s 6. A convenient notation to
identify this random variable is x N(35, 6).

40
6.4 Notation
• z-score used throughout statistics in a variety
of ways
• Need convenient notation to indicate the area
under the standard normal distribution
• z(a) is the algebraic name, for the z-score
(point on the z axis) such that there is a of the
area (probability) to the right of z(a)

41
Illustrations
42
Example
• Example Find the numerical value of z(0.10)

z(0.10) 1.28
43
Example
• Example Find the numerical value of z(0.80)
• Use Table 3 look for an area as close as
possible to 0.3000
• z(0.80) -0.84

44
Notes
• The values of z that will be used regularly come
from one of the following situations

1. The z-score such that there is a specified
area in one tail of the normal distribution
2. The z-scores that bound a specified middle
proportion of the normal distribution
45
Example
• Example Find the numerical value of z(0.99)
• Because of the symmetrical nature of the normal
distribution, z(0.99) -z(0.01)

46
Example
• Example Find the z-scores that bound the middle
0.99 of the normal distribution

47
6.5 Normal Approximation of the Binomial
• Recall the binomial distribution is a
probability distribution of the discrete random
variable x, the number of successes observed in n
repeated independent trials
• Binomial probabilities can be reasonably
estimated by using the normal probability
distribution

48
Background Histogram
• Background Consider the distribution of the
binomial variable x when n 20 and p 0.5
• Histogram

The histogram may be approximated by a normal
curve
49
Notes
• The normal curve has mean and standard deviation
from the binomial distribution
• Can approximate the area of the rectangles with
the area under the normal curve
• The approximation becomes more accurate as n
becomes larger

50
Two Problems
• 1. As p moves away from 0.5, the binomial
distribution is less symmetric, less
normal-looking

Solution The normal distribution provides a
reasonable approximation to a binomial
probability distribution whenever the values of
np and n(1 - p) both equal or exceed 5
2. The binomial distribution is discrete, and the
normal distribution is continuous
Solution Use the continuity correction factor.
Add or subtract 0.5 to account for the width of
each rectangle.
51
Example
• Example Research indicates 40 of all students
entering a certain university withdraw from a
course during their first year. What is the
probability that fewer than 650 of this years
entering class of 1800 will withdraw from a
class?

52
Solution
• Use the normal approximation method

53
Random Number Generation
• With each rand execution, the TI-84 Plus
generates the same random-number sequence
• for a given seed value. The TI-84 Plus
factory-set seed value for rand is 0. To generate
a
• different random-number sequence, store any
nonzero seed value to rand. To restore
• the factory-set seed value, store 0 to rand or
reset the defaults (Chapter 18).
• Note The seed value also affects randInt(,
randNorm(, and randBin( instructions.