The Normal Probability Distribution

1

• The Normal Probability
  Distribution

Chapter
2
THIS CHAPTERS GOALS

• TO LIST THE CHARACTERISTICS OF THE NORMAL
  DISTRIBUTION.
• TO DEFINE AND CALCULATE Z VALUES.
• TO DETERMINE PROBABILITIES ASSOCIATED WITH THE
  STANDARD NORMAL DISTRIBUTION.
• TO USE THE NORMAL DISTRIBUTION TO APPROXIMATE THE
  BINOMIAL DISTRIBUTION.

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CHARACTERISTICS OF A NORMAL PROBABILITY
DISTRIBUTION

• The normal curve is bell-shaped and has a single
  peak at the exact center of the distribution.
• The arithmetic mean, median, and mode of the
  distribution are equal and located at the peak.
• Half the area under the curve is above this
  center point, and the other half is below it.
• The normal probability distribution is
  symmetrical about its mean.
• It is asymptotic - the curve gets closer and
  closer to the x-axis but never actually touches
  it.

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CHARACTERISTICS OF A NORMAL DISTRIBUTION
Normal curve is symmetrical - two halves
identical -
Tail
Tail
0.5
0.5
Theoretically, curve extends to - infinity
Theoretically, curve extends to infinity
Mean, median, and mode are equal
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Normal Distributions with Equal Means but
Different Standard Deviations.
s 3.1 s 3.9 s 5.0
m 20
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Normal Probability Distributions with Different
Means and Standard Deviations.
m 5, s 3 m 9, s 6 m 14, s 10
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THE STANDARD NORMAL PROBABILITY DISTRIBUTION

• A normal distribution with a mean of 0 and a
  standard deviation of 1 is called the standard
  normal distribution.
• z value The distance between a selected value,
  designated X, and the population mean m, divided
  by the population standard deviation, s.
• The z-value is the number of standard deviations
  X is from the mean.

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EXAMPLE

• The monthly incomes of recent MBA graduates in a
  large corporation are normally distributed with a
  mean of 2,000 and a standard deviation of 200.
  What is the z value for an income X of 2,200?
  1,700?
• For X 2,200 and since z (X - m)/s, then
  z (2,200
  - 2,000)/200 1.
• A z value of 1 indicates that the value of 2,200
  is 1 standard deviation above the mean of 2,000.

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EXAMPLE (continued)

• For X 1,700 and since z (X - m)/s, then
  z (1,700 - 2,000)/200 -1.5.
• A z value of -1.5 indicates that the value of
  2,200 is 1.5 standard deviation below the mean
  of 2,000.
• How might a corporation use this type of
  information?

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AREAS UNDER THE NORMAL CURVE

• About 68 percent of the area under the normal
  curve is within plus one and minus one standard
  deviation of the mean. This can be written as
  m 1s.
• About 95 percent of the area under the normal
  curve is within plus and minus two standard
  deviations of the mean, written m 2s.
• Practically all (99.74 percent) of the area under
  the normal curve is within three standard
  deviations of the mean, written m 3s.

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Between 1. 68.26 2. 95.44 3. 99.97
m
m1s
m2s
m3s
m-1s
m-2s
m3s
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P(z)?

• A typical need is to determine the probability of
  a z-value being greater than or less than some
  value.
• Tabular Lookup (Appendix D, page 474)
• EXCEL Function NORMSDIST(z)

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EXAMPLE

• The daily water usage per person in Toledo, Ohio
  is normally distributed with a mean of 20 gallons
  and a standard deviation of 5 gallons.
• About 68 of the daily water usage per person in
  Toledo lies between what two values?

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EXAMPLE

• The daily water usage per person in Toledo (X),
  Ohio is normally distributed with a mean of 20
  gallons and a standard deviation of 5 gallons.
• What is the probability that a person selected at
  random will use less than 20 gallons per day?
• What is the probability that a person selected at
  random will use more than 20 gallons per day?

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EXAMPLE (continued)

• What percent uses between 20 and 24 gallons?
• The z value associated with X 20 is z 0 and
  with X 24, z (24 - 20)/5 0.8
  P(20 lt X lt 24) P(0 lt z lt 0.8)
  0.288128.81
• What percent uses between 16 and 20 gallons?

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P(0 lt z lt 0.8) 0.2881
0.8
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EXAMPLE (continued)

• What is the probability that a person selected at
  random uses more than 28 gallons?

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P(z gt 1.6) 0.5 - 0.4452 0.0048
Area 0.4452
z
1.6
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EXAMPLE (continued)

• What percent uses between 18 and 26 gallons?

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Total area 0.1554 0.3849 0.5403
Area 0.1554
Area 0.3849
z
- .4
1.2
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EXAMPLE (continued)

• How many gallons or more do the top 10 of the
  population use?
• Let X be the least amount. Then we need to find
  Y such that P(X ³ Y) 0.1 To find the
  corresponding z value look in the body of the
  table for (0.5 - 0.1) 0.4. The corresponding z
  value is 1.28 Thus we have (Y - 20)/5 1.28,
  from which Y 26.4. That is, 10 of the
  population will be using at least 26.4 gallons
  daily.

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(Y - 20)/5 1.28
Y 26.4
0.4
0.1
z
1.28
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EXAMPLE

• A professor has determined that the final
  averages in his statistics course is normally
  distributed with a mean of 72 and a standard
  deviation of 5. He decides to assign his grades
  for his current course such that the top 15 of
  the students receive an A. What is the lowest
  average a student must receive to earn an A?

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(Y - 72)/5 1.04
Y 77.2
0.35
0.15
z
1.04
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EXAMPLE

• The amount of tip the waiters in an exclusive
  restaurant receive per shift is normally
  distributed with a mean of 80 and a standard
  deviation of 10. A waiter feels he has provided
  poor service if his total tip for the shift is
  less than 65. Based on his theory, what is the
  probability that he has provided poor service?

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Area 0.5 - 0.4332 0.0668
Area 0.4332
z
- 1.5
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THE NORMAL APPROXIMATION TO THE BINOMIAL

• Using the normal distribution (a continuous
  distribution) as a substitute for a binomial
  distribution (a discrete distribution) for large
  values of n seems reasonable because as n
  increases, a binomial distribution gets closer
  and closer to a normal distribution.
• When to use the normal approximation?
• The normal probability distribution is generally
  deemed a good approximation to the binomial
  probability distribution when np and n(1 - p) are
  both greater than 5.

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Binomial Distribution with n 3 and p 0.5.
P(r)
0.5
0.4
0.3
0.25
r
0
1
2
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Binomial Distribution with n 5 and p 0.5.
P(r)
r
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Binomial Distribution with n 20 and p 0.5.
P(r)
Observe the Normal shape.
r
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THE NORMAL APPROXIMATION (continued)

• Recall for the binomial experiment
• There are only two mutually exclusive outcomes
  (success or failure) on each trial.
• A binomial distribution results from counting the
  number of successes.
• Each trial is independent.
• The probability p is fixed from trial to trial,
  and the number of trials n is also fixed.

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CONTINUITY CORRECTION FACTOR

• The value 0.5 subtracted or added, depending on
  the problem, to a selected value when a binomial
  probability distribution, which is a discrete
  probability distribution, is being approximated
  by a continuous probability distribution--the
  normal distribution.
• The basic concept is that a slice of the normal
  curve from x-0.5 to x0.5 is approximately equal
  to P(x).

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EXAMPLE

• A recent study by a marketing research firm
  showed that 15 of the homes had a video recorder
  for recording TV programs. A sample of 200 homes
  is obtained. (Let X be the number of homes).
• Of the 200 homes sampled how many would you
  expect to have video recorders?
• m np (0.15)(200) 30 n(1 - p) 170
• What is the variance?
• s2 np(1 - p) (30)(1- 0.15) 25.5

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EXAMPLE (continued)

• What is the standard deviation?
• s Ö(25.5) 5.0498.
• What is the probability that less than 40 homes
  in the sample have video recorders?
• We need P(X lt 40) P(X 39). So, using the
  normal approximation, P(X 39.5)
• Pz (39.5 - 30)/5.0498 P(z
  1.8812)
• P(z 1.88) 0.5 0.4699 0.9699
• Why did I use 39.5 ? ...
• How would you calculate P(X39) ?

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P(z 1.88) 0.5 0.4699 0.9699
0.5
0.4699
z
1.88
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EXAMPLE (continued)

• What is the probability that more than 24 homes
  in the sample have video recorders?

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P(z ³ -1.09) 0.5 0.3621 0.8621.
0.5
0.3621
z
-1.09
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EXAMPLE (continued)

• What is the probability that exactly 40 homes in
  the sample have video recorders?

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P(1.88 z 2.08) 0.4812 - 0.4699 0.0113
1.88
2.08
0.4699
0.4812
z