Chapter 7 Section 3

1
Chapter 7Section 3

• Applications of the
• Normal Distribution

2
Chapter 7 Section 3

• Learning objectives
• Find and interpret the area under a normal curve
• Find the value of a normal random variable

3
Chapter 7 Section 3

• Learning objectives
• Find and interpret the area under a normal curve
• Find the value of a normal random variable

4
Chapter 7 Section 3

• We want to calculate probabilities and values for
  general normal probability distributions
• We can relate these problems to calculations for
  the standard normal in the previous section

5
Chapter 7 Section 3

• For a general normal random variable X with mean
  µ and standard deviation s, the variable
• has a standard normal probability distribution
• We can use this relationship to perform
  calculations for X

6
Chapter 7 Section 3

• Values of X ?? Values of Z
• If x is a value for X, then
• is a value for Z
• This is a very useful relationship

7
Chapter 7 Section 3

• For example, if
• µ 3
• s 2

• For example, if
• µ 3
• s 2
• then a value of x 4 for X corresponds to

• For example, if
• µ 3
• s 2
• then a value of x 4 for X corresponds to
• a value of z 0.5 for Z

8
Chapter 7 Section 3

• Because of this relationship
• Values of X ?? Values of Z
• then
• P(X lt x) P(Z lt z)

• Because of this relationship
• Values of X ?? Values of Z
• then
• P(X lt x) P(Z lt z)
• To find P(X lt x) for a general normal random
  variable, we could calculate P(Z lt z) for the
  standard normal random variable

9
Chapter 7 Section 3

• This relationship lets us compute all the
  different types of probabilities
• Probabilities for X are directly related to
  probabilities for Z using the (X µ) / s
  relationship

10
Chapter 7 Section 3

• A different way to illustrate this relationship

X
a µ b
Z
11
Chapter 7 Section 3

• With this relationship, the following method can
  be used to compute areas for a general normal
  random variable X

• With this relationship, the following method can
  be used to compute areas for a general normal
  random variable X
• Shade the desired area to be computed for X

• With this relationship, the following method can
  be used to compute areas for a general normal
  random variable X
• Shade the desired area to be computed for X
• Convert all values of X to Z-scores using

• With this relationship, the following method can
  be used to compute areas for a general normal
  random variable X
• Shade the desired area to be computed for X
• Convert all values of X to Z-scores using
• Solve the problem for the standard normal Z

• With this relationship, the following method can
  be used to compute areas for a general normal
  random variable X
• Shade the desired area to be computed for X
• Convert all values of X to Z-scores using
• Solve the problem for the standard normal Z
• The answer will be the same for the general
  normal X

12
Chapter 7 Section 3

• Examples
• For a general random variable X with
• µ 3
• s 2
• calculate P(X lt 6)

• Examples
• For a general random variable X with
• µ 3
• s 2
• calculate P(X lt 6)
• This corresponds to
• so P(X lt 6) P(Z lt 1.5) 0.9332

13
Chapter 7 Section 3

• Examples
• For a general random variable X with
• µ 2
• s 4
• calculate P(X gt 3)

• Examples
• For a general random variable X with
• µ 2
• s 4
• calculate P(X gt 3)
• This corresponds to
• so P(X gt 3) P(Z gt 0.25) 0.5987

14
Chapter 7 Section 3

• Examples
• For a general random variable X with
• µ 6
• s 4
• calculate P(4 lt X lt 11)

• Examples
• For a general random variable X with
• µ 6
• s 4
• calculate P(4 lt X lt 11)
• This corresponds to
• so P(4 lt X lt 11) P( 0.5 lt Z lt 1.25) 0.5858

15
Chapter 7 Section 3

• Technology often has direct calculations for the
  general normal probability distribution

• Technology often has direct calculations for the
  general normal probability distribution
• Excel
• The function NORMDIST (instead of NORMSDIST)
  calculates general normal probabilities

• Technology often has direct calculations for the
  general normal probability distribution
• Excel
• The function NORMDIST (instead of NORMSDIST)
  calculates general normal probabilities
• StatCrunch
• Entering the values for the mean and standard
  deviation into the window turns the standard into
  a general normal calculator

16
Chapter 7 Section 3

• Learning objectives
• Find and interpret the area under a normal curve
• Find the value of a normal random variable

17
Chapter 7 Section 3

• The inverse of the relationship
• is the relationship
• With this, we can compute value problems for the
  general normal probability distribution

18
Chapter 7 Section 3

• The following method can be used to compute
  values for a general normal random variable X

• The following method can be used to compute
  values for a general normal random variable X
• Shade the desired area to be computed for X

• The following method can be used to compute
  values for a general normal random variable X
• Shade the desired area to be computed for X
• Find the Z-scores for the same probability problem

• The following method can be used to compute
  values for a general normal random variable X
• Shade the desired area to be computed for X
• Find the Z-scores for the same probability
  problem
• Convert all the Z-scores to X using

• The following method can be used to compute
  values for a general normal random variable X
• Shade the desired area to be computed for X
• Find the Z-scores for the same probability
  problem
• Convert all the Z-scores to X using
• This is the answer for the original problem

19
Chapter 7 Section 3

• Examples
• For a general random variable X with
• µ 3
• s 2
• find the value x such that P(X lt x) 0.3

• Examples
• For a general random variable X with
• µ 3
• s 2
• find the value x such that P(X lt x) 0.3
• Since P(Z lt 0.5244) 0.3, we calculate
• so P(X lt 1.95) P(Z lt 0.5244) 0.3

20
Chapter 7 Section 3

• Examples
• For a general random variable X with
• µ 2
• s 4
• find the value x such that P(X gt x) 0.2

• Examples
• For a general random variable X with
• µ 2
• s 4
• find the value x such that P(X gt x) 0.2
• Since P(Z gt 0.8416) 0.2, we calculate
• so P(X gt 1.37) P(Z gt 0.8416) 0.2

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Chapter 7 Section 3

• Examples
• We know that z.05 1.28, so
• P(1.28 lt Z lt 1.28) 0.90

• Examples
• We know that z.05 1.28, so
• P(1.28 lt Z lt 1.28) 0.90
• Thus for a general random variable X with
• µ 6
• s 4

• Examples
• We know that z.05 1.28, so
• P(1.28 lt Z lt 1.28) 0.90
• Thus for a general random variable X with
• µ 6
• s 4
• so P(0.58 lt X lt 12.58) 0.90

22
Chapter 7 Section 3

• Technology often has direct calculations for the
  general normal probability distribution

• Technology often has direct calculations for the
  general normal probability distribution
• Excel
• The function NORMINV (instead of NORMSINV)
  calculates general normal probabilities

• Technology often has direct calculations for the
  general normal probability distribution
• Excel
• The function NORMINV (instead of NORMSINV)
  calculates general normal probabilities
• StatCrunch
• Entering the values for the mean and standard
  deviation into the window turns the standard into
  a general normal calculator

23
Summary Chapter 7 Section 3

• We can perform calculations for general normal
  probability distributions based on calculations
  for the standard normal probability distribution
• For tables, and for interpretation, converting
  values to Z-scores can be used
• For technology, often the parameters of the
  general normal probability distribution can be
  entered directly into a routine

24
Example Chapter 7 Section 3

• The combined (verbal quantitative reasoning)
  score on the GRE is normally distributed with
  mean 1066 and standard deviation 191. (Source
  www.ets.org/Media/Tests/GRE/pdf/01210.pdf.) The
  Department of Psychology at Columbia University
  in New York requires a minimum combined score of
  1200 for admission to their doctoral program.
  (Source www.columbia.edu/cu/gsas/departments/psyc
  hology/department.html.)
• a. What proportion of combined GRE scores can be
  expected to be under 1100? (0.5706)
• b. What proportion of combined GRE scores can be
  expected to be over 1100? (0.4294)
• c. What proportion of combined GRE scores can be
  expected to be between 950 and 1000? (0.0930)
• d. What is the probability that a randomly
  selected student will score over 1200 points?
  (0.2415)
• e. What is the probability that a randomly
  selected student will score less than 1066
  points? (0.5000)

25
Example Chapter 7 Section 3

• The combined (verbal quantitative reasoning)
  score on the GRE is normally distributed with
  mean 1066 and standard deviation 191. (Source
  www.ets.org/Media/Tests/GRE/pdf/01210.pdf.) The
  Department of Psychology at Columbia University
  in New York requires a minimum combined score of
  1200 for admission to their doctoral program.
  (Source www.columbia.edu/cu/gsas/departments/psyc
  hology/department.html.)
• f. What is the percentile rank of a student who
  earns a combined GRE score of 1300? (89th
  percentile)
• g. What is the percentile rank of a student who
  earns a combined GRE score of 1000? (36th
  percentile)
• h. Determine the 70th percentile of combined GRE
  scores. (1166)
• i. Determine the combined GRE scores that make up
  the middle 95 of all scores. (692 to 1440)
• j. Compare the results in part (i) to the values
  given by the Empirical Rule. (684 to 1448 they
  are very close, since the Empirical Rule is based
  on the normal distribution.)

26
Example Chapter 7 Section 3

• The diameters of ball bearings produced at a
  factory are approximately normally distributed.
  Suppose the mean diameter is 1.002 centimeters
  (cm) and the standard deviation is 0.006 cm. The
  product specifications require that the diameter
  of each ball bearing be between 0.980 and 1.020
  cm.
• a. What proportion of ball bearings can be
  expected to have a diameter under 1.020 cm?
  (0.9987)
• b. What proportion of ball bearings can be
  expected to have a diameter over 1.020 cm?
  (0.0013)
• c. What proportion of ball bearings can be
  expected to have a diameter between 0.980 and
  1.020 cm? That is, what proportion of ball
  bearings can be expected to meet the
  specifications? (0.9986)

27
Example Chapter 7 Section 3

• The diameters of ball bearings produced at a
  factory are approximately normally distributed.
  Suppose the mean diameter is 1.002 centimeters
  (cm) and the standard deviation is 0.006 cm. The
  product specifications require that the diameter
  of each ball bearing be between 0.980 and 1.020
  cm.
• d. What is the probability that the diameter of a
  randomly selected ball bearing will be over 1.000
  cm? (0.6306)
• e. What is the probability that the diameter of a
  randomly selected ball bearing will be under
  0.995 cm? (0.1217)
• f. What is the percentile rank of a ball bearing
  that has a diameter of 0.991 cm? (3rd percentile)
  
• g. What is the percentile rank of a ball bearing
  that has a diameter of 1.011 cm? (93rd
  percentile)
• h. Determine the 10th percentile of the diameters
  of ball bearings. (0.994 cm)
• i. Determine the diameters of ball bearings that
  make up the middle 99 of all diameters. (0.987
  to 1.017 cm)