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The concept of binomial distributions and their relationship to normal distributions. It covers the binomial probability theorem, the concept of a probability function and distribution, and the graphical representation of normal distributions. The document also includes examples and exercises to help students understand these concepts.
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Typology: Exercises
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922 Series and Combinations
Chapter 13
BIG IDEA As the number of trials of a binomial experiment increases, the graphs of the probabilities of each event and the relative frequencies of each event approaches a distribution called a normal distribution.
Let P ( n ) = the probability of n heads in 5 tosses of a fair coin. Then the domain of P is {0, 1, 2, 3, 4, 5}. By the Binomial Probability Theorem, and because the probability of heads = the probability of tails = _^12 ,
⎝ n ⎠ =^
_^1 32
⎝ n ⎠.
a. Copy and complete the table shown below. b. Graph and label the coordinates of the six points (n, P(n)). Solution a. Use the formula for P(n) given above to fill in the second row. n = Number of Heads 0 1 2 3 4 5 P(n) = Probability of n Heads in 5 Tosses of a Fair Coin
5
32
5
? (^)????
b. The points are graphed below. Fill in the missing coordinates with values from the table in Part a.
1 8
0 1 2 3 4 5
1 32
1 4
(0, )
(1,?^ ) (4,^ ?)
(5, ?)
(2,?^ ) (3, ?)
Probability = P ( n )
n = Number of Heads
P ( n ) = 32
5 n
⎛⎪ ⎝
⎞⎥ ⎠
1 8
0 1 2 3 4 5
1 32
1 4
(0, )
(1,?^ ) (4,^ ?)
(5, ?)
(2,?^ ) (3, ?)
Probability = P ( n )
n = Number of Heads
P ( n ) = 32
5 n
⎛⎪ ⎝
⎞⎥ ⎠
Mental Math
Tell whether each graph is the graph of a function. a. the image of y = sin x under R 90 b. the image of y = cos x under r (^) x-axis c. the image of x = 12 under T2, 8 d. the image of x = 12 under R 3
probability function, probability distribution binomial probability distribution, binomial distribution normal distribution normal curve standard normal curve standardized scores
Binomial and Normal Distributions 923
Lesson 13-
P is a probability function. A probability function, or probability distribution, is a function that maps a set of events onto their probabilities. Because the function P results from calculations of binomial probabilities, it is called a binomial probability distribution, or simply a binomial distribution.
If a fair coin is tossed 10 times, the possible numbers of heads are 0, 1, 2, ..., 10, so there are 11 points in the graph of the corresponding probability function. Again, by the Binomial Probability Theorem with equally likely events, the probability P ( x ) of tossing x heads is given by
P ( x ) = (^) ( _^12 )
⎝ x ⎠ =^
__^1 1024
⎝ x ⎠.
The 11 probabilities are easy to calculate because the numerators in the fractions are the numbers in the 10th row of Pascal’s triangle. That is, they are binomial coefficients.
n = Number of Heads 0 1 2 3 4 5 6 7 8 9 10
P ( x ) = Probability of x Heads
__ 1 1024 ≈ 0.
____^10 1024 ≈ 0.
____^45 1024 ≈ 0.
_^120 1024 ≈ 0.
_^210 1024 ≈ 0.
_^252 1024 ≈ 0.
_^210 1024 ≈ 0.
_^120 1024 ≈ 0.
_^45 1024 ≈ 0.
_^10 1024 ≈ 0.
_^1 1024 ≈ 0.
The binomial distribution in the table is graphed at the right. Closely examine this 11-point graph of
P ( x ) = _ 10241
⎝ x ⎠, along with the table of values. The individual probabilities are all less than 1 __ 4.
Notice how unlikely it is to get 0 heads or 10 heads in a row. (The probability for each is less than _ 10001 .)
Even for 9 heads in 10 tosses, the probability is less than _ 10001. Like the graph of the 6-point probability
function P ( n ) = _ 321 ⎛⎝^ n^5 ⎞⎠ on the previous page, this
11-point graph has a vertical line of symmetry.
As the number of tosses of a fair coin is increased, the points on the graph more closely outline a curve shaped like a bell. On the next page, this bell-shaped curve is positioned so that it is reflection-symmetric to the y -axis and its equation is simplest.
Its equation is y = _^1 √ 2 π
_– x^2
0 2 4 6 8 10 x = (^) Number of Heads
Probability =^ P ( x )
P : x → 1024
10 x
⎛⎪ ⎝
⎞⎥ 1 ⎠ 4
256 = 1024
1 8
128 = 1024
3 16
192 = 1024
1 16
64 = 1024
=
=
=
=
**0.
0.**
0 2 4 6 8 10 x = (^) Number of Heads
Probability =^ P ( x )
P : x → 1024
10 x
⎛⎪ ⎝
⎞⎥ 1 ⎠ 4
256 = 1024
1 8
128 = 1024
3 16
192 = 1024
1 16
64 = 1024
=
=
=
=
**0.
0.**
Binomial and Normal Distributions 925
Lesson 13-
Normal curves are often good mathematical models for the distribution of scores on an exam. The graph at the right shows an actual distribution of scores on a 40-question test given to 209 geometry students. (It was a hard test!) A possible corresponding normal curve is shown.
On some tests, scores are standardized. This means that a person’s score is not the number of correct answers, but is converted so that it lies in a normal distribution with a predetermined mean and standard deviation. Standardized tests make it easy to evaluate an individual score relative to the mean, but not to other individual scores.
SAT scores are standardized to a historic mean of 500 and a standard deviation of 100. What percent of the scores are expected to be between 400 and 700?
Solution Find out how many standard deviations each score, 400 and 700, is from the mean, 500. 500 - 400 = 100 and 700 - 500 = 200 So, a score of 400 is 1 standard deviation below the mean and a score of 700 is 2 standard deviations above the mean. Refer to the Normal Distribution Percentages graph on the previous page. The percent of scores between μ - 1 σ and μ + 2 σ is 34.1% + 34.1% + 13.6% = 81.8%. About 81.8% of SAT scores are expected to be between 400 and 700.
The normal distribution is an appropriate topic with which to end this book, because it involves so many of the ideas you have studied in it. The distribution is a function. Its equation involves squares, square roots, π, e , and negative exponents. Its graph is the composite of a translation and scale change image of the curve with equation
y = _^1 √ 2 π
that is used on tests that help to determine which colleges some people will attend. It shows how interrelated and important the ideas of mathematics are.
0 4 8 12 16 20 24 28 32 36
4
6
8
10
12
14
20 16
18
2
Number of Students with Score x
x = Score
Mean = 19. Standard Deviation = 5.
0 4 8 12 16 20 24 28 32 36
4
6
8
10
12
14
20 16
18
2
Number of Students with Score x
x = Score
Mean = 19. Standard Deviation = 5.
926 Series and Combinations
Chapter 13
a. What kind of function is P? b. Find P (4) and describe what it could represent.