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AMS7: WEEK 5. CLASS 1
Sampling distributions and estimators.
Central Limit Theorem
Normal Approximation to the Binomial Distribution
Monday April 27th, 2015
Example of application of the Central Limit
Theorem
• Assume that men’s weights are normally distributed with a
given mean ߤ= 172 lb. and a standard deviation given by ߪ
=29 lb.
1) If one man is selected at random, find the probability that
his weight is less than 167 lb. Random variable X is
mean’s weight
a) Draw a picture:
167
172
Weight
Example (Cont.)
b) Calculate z Score
‫=ݖ‬
௫ିఓ
ఙ
=
ଵ଺଻ିଵ଻ଶ
=
ଶଽ
-0.17
c) Area= P(Weight < 167)=P(z<-0.17)=0.4325
2) If 36 men are randomly selected, find the probability that
they have a mean weight less than 167 lb.
Now we have a sample of 36 men
Example (Cont.)
a) Use Central Limit Theorem:
has a normal distribution with mean ௑ത = and a standard deviation ௑ത =
ఙ
௡
=
ଶଽ
ଷ଺
=
ଶଽ
଺
= 4.83
b) Find the z Score for :
‫=ݖ‬
௫ିఓ೉
ഥ
ఙ೉
ഥ
=
ଵ଺଻ିଵ଻ଶ
=
ସ.଼ଷ
3) Find the area below the z Score:
Area=P(Mean weight < 167)=P(z<-1.04)=0.1492
-1.04
Normal approximation to the binomial
• When working with a binomial distribution if np ≥ 5 and
nq ≥ 5 the binomial random variable has a probability
distribution that can be approximated by a normal
distribution, with mean and standard deviation:
ߤ = n.p
࣌ = ࢔. ࢖. ࢗ
We must verify that it is reasonable to approximate the
binomial distribution by the normal distribution
Normal approximation to the binomial
(cont.)
EXAMPLE: Suppose that X is a random variable with a binomial
distribution (Example: Number of left-handed in a Statistics
class)
1) With n=200, p=0.5
• q=1-p=0.5
• np= 200⨯0.5=100 (np ≥ 5)
• qn= 200⨯0.5=100 (nq ≥ 5)
2) Then we can calculate:
ߤ = n.p= 200⨯0.5=100
࣌ = ࢔. ࢖. ࢗ= ૛૙૙ × ૙. ૞ × ૙. ૞= 7.07
Normal approximation to the binomial
(cont.)
c) Find the probability that X is less than 120 (P(X<120).
We can use the normal approximation (it is easier!) but we
need to have a continuity correction because the binomial
distribution is discrete and the normal distribution is
continuous.
Number 120 will be represented by the interval (120-0.5,
120+0.5)= (119.5,120.5) in the normal curve.
Normal approximation to the binomial
(cont.)
100
(119.5 120.5)
120
Normal approximation to the binomial
(cont.)
• We want the z Score:
‫=ݖ‬
௫ିఓ
ఙ
=
ଵଵଽ.ହିଵ଴଴
=
଻.଴଻
2.76
• P(z<2.76)=P(X<120)
• P(x≤119.5)= P(z≤ 2.76)=0.9971 (Using the continuity
correction)
Example
• Normal approximation to the Binomial distribution
Estimate the probability of at least passing a true/false test
of 100 questions if 60% (or 60 correct answers) is the
minimum passing grade and all responses are random
guesses. Is that probability high enough to risk passing
using random guesses??
• In this problem the random variable is X= the number of
right answers in 100 questions. X is a Binomial random
variable:
n=100
p=0.5 (probability of getting a right answer)
Solution
• We want to calculate P(X≥60)= Probability of at least
passing the exam.
1) Since n.p = 50 ≥ 5 and n.q=50 ≥5 we can use the
normal approximation to the binomial model.
2) Calculate mean and standard deviation:
ߤ = n.p= 100⨯0.5=50
࣌ = ࢔. ࢖. ࢗ= ૚૙૙ × ૙. ૞ × ૙. ૞ = ૛૞ = ૞
3) Calculate the z Score using a correction if necessary:
‫ ݔ‬− ߤ 59.5 − 50 9.5
‫=ݖ‬
=
=
= 1.9
ߪ
5
5
Solution (Cont.)
4) P(X≥60)=P(Z>1.9)=1-0.9713=0.0287 <0.05
Conclusion: It is unusual to get a score of at least 60 by
guessing!!!
Corrections for continuity
• At least 60
59.5
• More than 60
60.5
• At most 60
60.5
• Fewer than 60
59.5
• Exactly 60
59.5 60.5