Download Normal approx. to binomial

Normal Approximation
to the Binomial
Business Statistics
Plan for Today
• The Binomial Distribution
• Approximating the Binomial Distribution by
the Normal Distribution
• Examples
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The Binomial Distribution
• The binomial distribution is a discrete probability
distribution (i.e. the random variable x has only a
discrete or a finite set of possible values).
• The binomial distribution requires n
• repeated
• independent
• identical
trials. Each trial is either success or failure.
Probability of success = p,
probability of failure = q = 1 – p .
The random variable x = the number of successes.
Example: tossing a coin
• A fair coin is tossed or flipped n times in an
identical independent manner.
• The random variable x will count
the number of “tails”.
• Success = “tails”.
• Failure = “heads”.
• p=½
• q=1–½=½
• x can take values from 0 to n
• What is the mean value of x ?
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Example: rolling a die
• A fair die is rolled n times in an identical,
independent manner.
• Declare, for example, that the
random variable x counts the
number of times a “6” is rolled.
• Success = “6”
• Failure = “1”, “2”, “3”, “4”, or “5”.
• p = 1/6
and
q = 5/6 = 1 – 1/6
• Again, the variable x can take values from 0 to n.
• What is the mean value of x ?
Example: blood type B
• It is known that 9% of Canadians have blood type B.
• Suppose that n Canadians are chosen at random.
• x – the number of Canadians among
those with blood type B.
• Success = type B
• Failure = type O, A, or AB
• p = 0.09 and q = 1 – 0.09 = 0.91
• The random variable x can take values
from 0 to n. What is its mean?
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Example: smokers
• 21% of adults in Quebec are smokers
• 40 adult Quebecois are chosen at random
• x – the number of smokers
among these
• Success = smoker (?!)
• Failure = non-smoker
• p = 0.21 and q = 1 – 0.21 = 0.79
• The random variable x takes values
between 0 and 40.
• Mean 𝜇 = 8.4. Why?
The Binomial Distribution
• The random variable x (the number of
successes) has the following measures:
• The mean
𝝁=𝒏∙𝒑
• The variance
𝝈𝟐 = 𝒏 ∙ 𝒑 ∙ 𝒒
• The standard deviation
𝝈=
𝝈𝟐 = 𝒏𝒑𝒒
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Approximating the Binomial Distribution
by the Normal Distribution
• A consequence of the CLT: a Binomial
Distribution 𝐵(𝑛, 𝑝) with the number of
trials n and the probability of success p can
be approximated by the normal distribution
𝑁(𝜇, 𝜎) if the following two conditions hold:
𝑛 ∙ 𝑝 > 5 and 𝑛 ∙ 𝑞 > 5
• Always check these conditions before
applying approximation.
Continuity Correction Factor
• Instead of x we will use x – 0.5 or x + 0.5 depending
on the contextual question.
• “More than 30”: x = 30.5
• “At least 30”: x = 29.5
• “Less than 46”: x = 45.5
• “At most 46”: x = 46.5
• “Between 34 and 37 (inclusively)”:
x1 = 33.5 and x2 = 37.5
• “Exactly 27”: x1 = 26.5 and x2 = 27.5
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Example: tossing a coin
• A fair coin is tossed 16 times. What is the probability
that the number of “heads” is at least 11?
• p = 0.5, q = 0.5, 𝜇 = 𝑛𝑝 = 8, 𝜎 = 𝑛𝑝𝑞 = 2
• Checking conditions: 𝑛𝑝 > 5, 𝑛𝑞 > 5 ? Yes.
• Correction factor: x = 11 – 0.5 = 10.5
• The z-score: 𝑧 =
10.5−8
2
= 1.25
• The required probability:
𝑃 𝑧 > 1.25 = 0.5 − 𝑇 1.25 = 0.5 − 0.3944
= 0.1056 or 10.56%
Example: rolling a die
• A fair die is rolled 33 times. Find the probability that
the number “4” was rolled less than 8 times.
• p = 1/6, q = 5/6, 𝜇 = 𝑛𝑝 = 5.5, 𝜎 = 𝑛𝑝𝑞 = 2.14
• Checking conditions: 𝑛𝑝 > 5, 𝑛𝑞 > 5 ? Yes.
• Correction factor: x = 8 – 0.5 = 7.5
• The z-score: 𝑧 =
7.5−5.5
2.14
= 0.93
• The required probability:
𝑃 𝑧 > 0.93 = 0.5 + 𝑇 0.93 = 0.5 + 0.3238
= 0.8238 or 82.38%
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Example: blood type B
• 90 Canadians are chosen at random. Find the
probability that between 8 and 14 (inclusively) have
blood type B.
• p = 0.09, q = 0.91, 𝜇 = 𝑛𝑝 = 8.1, 𝜎 = 𝑛𝑝𝑞 = 2.71
• Checking conditions: 𝑛𝑝 > 5, 𝑛𝑞 > 5 ? Yes.
• Correction factors: x1 = 7.5 and x2 = 14.5
• The z-scores: 𝑧1 = −0.22 and 𝑧2 = 2.36
• The required probability:
𝑃 −0.22 < 𝑧 < 2.36 = 𝑇(0.22) + 𝑇 2.36 =
0.0871 + 0.4909 = 0.5780 or 57.80%
Example: smokers
• 30 Quebecois are chosen at random. Find the
probability that exactly 5 of those are smokers.
• p = 0.21, q = 0.79, 𝜇 = 𝑛𝑝 = 6.3, 𝜎 = 𝑛𝑝𝑞 = 2.23
• Checking conditions: 𝑛𝑝 > 5, 𝑛𝑞 > 5 ? Yes.
• Correction factors: x1 = 4.5 and x2 = 5.5
• The z-scores: 𝑧1 = −0.81 and 𝑧2 = −0.36
• The required probability:
𝑃 −0.81 < 𝑧 < −0.36 = 𝑇 0.81 − 𝑇 0.36 =
0.2910 − 0.1406 = 0.1504 or 15.04%
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