Download Math160.Section6.5

Math 160 - Cooley
Intro to Statistics
OCC
Section 6.5 – Normal Approximation to the Binomial Distribution
Recall the formulas from Section 5.3:
Binomial Probability Formula :
n
P ( X  x)    p x q n  x , x = 0, 1, 2, …, n.
 x
Mean and Standard Deviation of a Binomial Random Variable:
  np and   npq
In this section, we can use the normal distribution to approximate a binomial distribution under certain
circumstances, just as we used a Poisson distribution to approximate a binomial distribution in Section 5.4.
To Approximate Binomial Probabilities by Normal-Curve Areas
Step 1:
Find n, the number of trials, and p, the success probability.
Step 2:
Continue only if both np and n(1  p) are 5 or greater.
Step 3:
Find  and  , using the formulas   np and   npq .
Step 4:
Make the correction for continuity, and find the required area under the normal curve with
parameters  and  .
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Math 160 - Cooley
Intro to Statistics
OCC
Section 6.5 – Normal Approximation to the Binomial Distribution
 Example:
A student is taking a true-false exam with 10 questions. Assume that the student guesses at all 10 questions.
a)
Determine the probability that the student gets either 7 or 8 questions correct.
Solution 
10 
10 
7
3
8
2
P(X = 7 or 8) = P(X = 7) + P(X = 8) =    0.5  0.5     0.5  0.5  0.1172  0.0439  0.1611
7
8
b)
Approximate the probability obtained in part (a) by an area under a suitable normal curve.
Solution 
In order to use the approximation, we need to check that both np and n(1  p) are 5 or greater.
So, np  10(0.5)  5 and n(1  p)  10(1  0.5)  5 , which are both 5 or greater.
Next, we need to find  and  . So,   np  10(0.5)  5 and   npq  10(0.5)(1  0.5)  1.58
Now, when making the correction for continuity, we will illustrate using the histogram below:
The histogram shows the binomial distribution for the problem. The highlighted bars (in light blue)
are for the question in part (a): P(X = 7 or 8) . The normal curve is shown overlapping (and
approximating) the histogram of the binomial distribution.
For our problem, using a normal distribution, P( X  7 or 8) is equivalent to P(7  X  8) . So, when
using the normal approximation, we must account for the correction for continuity. So, we need to
subtract 0.5 from the left limit and add 0.5 to the right limit, as shown in the figure. Thus,
P(7  X  8) is equivalent to P(6.5  X  8.5) , when using the normal approximation. Hence,
 6.5  5 x   8.5  5 
P(6.5  X  8.5)  P 


 P  0.95  z  2.22   0.1579

1.58 
 1.58
As you can see, the approximation using a normal distribution is 0.1579. When using an exact binomial
distribution, the probability was 0.1611. Thus, the approximation was a good one indeed.
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Math 160 - Cooley
Intro to Statistics
OCC
Section 6.5 – Normal Approximation to the Binomial Distribution
 Exercises:
1)
As reported in an issue of Weatherwise, according to the National Oceanic and Atmospheric
Administration, people at ballparks and playgrounds are in more danger of being struck by lightning
than are those on golf courses. What is the probability that, of 250 randomly selected lightning-induced
fatalities, the number occurring on golf courses is between 4 and 10 inclusive.
 Experiment:
2)
Each person in class is to flip a coin 20 times and record the outcomes below:
Flips 1 thru 20
Count the number of heads recorded: __________.
Now, record your number to your instructor. The entire class’s data will be displayed below:
Number of heads
in 20 flips
Tally
Frequency
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
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Math 160 - Cooley
Intro to Statistics
OCC
Section 6.5 – Normal Approximation to the Binomial Distribution
a)
Find the mean probabilityach
The Probability Distribution for 20 flips of a fair coin.
x
0
1
P(x)
 20 
P(X = 0) =   .50 .520 = .00000095
0
 20 
P(X = 1) =   .51 .519 = .0000191
1
11
12
P(x)
 20 
P(X = 11) =   .511 .59 = .16018
 11 
 20 
P(X = 12) =   .512 .56 = .12013
 12 
 20 
 .52 .518 = .0001812
2
13
P(X = 13) = 
 20 
 .53 .517 = .001087
3
14
P(X = 14) = 
 20 
 .54 .516 = .004621
4
15
P(X = 15) = 
16
P(X = 16) = 
2
P(X = 2) = 
3
P(X = 3) = 
4
P(X = 4) = 
5
P(X = 5) = 
6
x
 20 
 .55 .515 = .01479
5
 
 20 
P(X = 6) =   .56 .514 = .03696
6
17
 20 
 .513 .57 = .07393
 13 
 20 
 .514 .56 = .03696
 14 
 20 
 .515 .55 = .01479
 15 
 20 
 .516 .54 = .004621
16
 
 20 
P(X = 17) =   .517 .53 = .001087
 17 
 20 
 .57 .513 = .07393
7
 
18
P(X = 18) = 
 20 
 .58 .512 = .12013
8
 
19
P(X = 19) = 
 20 
 .59 .511 = .16018
9
 
20
P(X = 20) = 
7
P(X = 7) = 
8
P(X = 8) = 
9
P(X = 9) = 
10
P(X = 10) = 
 20 
 .518 .52 = .0001812
18
 
 20 
 .519 .51 = .0000191
19
 
 20 
 .520 .50 = .00000095
20
 
 20 
 .510 .510 = .17620
10
 
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