Chapter 6 ~ Normal Probability Distributions

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Chapter 6 Normal Probability Distributions
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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

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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

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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
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The Normal Probability Distribution
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Probabilities for a Normal Distribution
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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

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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

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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)

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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

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Table 3, Appendix B Entries

• The table contains the area under the standard
  normal curve between 0 and a specific value of z

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Example

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

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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

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Example

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

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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

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Example

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

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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

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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
  sketch is always helpful

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Example

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

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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

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Example

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

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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

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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?

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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.

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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

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Example

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

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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.

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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)

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Using the TI 83/84

• Now lets work the problems on page 279

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6.3 Applications of Normal Distributions

• Apply the techniques learned for the z
  distribution to all normal distributions

• Start with a probability question in terms
  ofx-values

• Convert, or transform, the question into an
  equivalent probability statement
  involvingz-values

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Standardization

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

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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
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Solution Continued
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Example, Part 2
2)
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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.

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Solution
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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.

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Solution
-
m
x


z


s
.

1
88
10
s
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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).

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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)

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Illustrations
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Example

• Example Find the numerical value of z(0.10)

z(0.10) 1.28
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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

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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
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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)

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Example

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

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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

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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
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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

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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.
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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?

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Solution

• Use the normal approximation method

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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.