Download 5.5 Distributions for Counts

5.5 Distributions for Counts
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1
Binomial Distributions for Sample Counts
Finding Binomial Probabilities
Binomial Mean and Standard Deviation
Binomial Formula
Binomial in Statistical Sampling
Binomial distributions for sample counts
Binomial distributions are models for some categorical variables, typically
counting the number of successes in a series of n independent trials
(think: # of Heads in 10 tosses of a fair coin)
The trials in a binomial experiment must meet these requirements:
– The total number of trials n is fixed in advance.
– The result of each trial can be put into just 1 of 2 categories:
“success” and “failure”.
– The n trials are independent of each other.
– Each trial has the same probability of “success,” that we call p.
We let X = the count of successes; in a binomial setting, X is a r.v. and
has a binomial distribution with parameters n and p, where n is the
number of trials and p is the probability of a “success” on any one trial.
The possible values of X are the whole numbers from 0 to n and for
many n and p, their probabilities are given in Table C. We may also
compute these probabilities using the TI-83 and JMP.
Binomial Mean and Standard
Deviation
If a count X has the binomial distribution based on n observations with
probability p of success, what is its mean µ? In general, the mean of a
binomial distribution should be µ = np. Here are the facts:
Mean and Standard Deviation of a Binomial Random Variable
If a count X has the binomial distribution with number of trials n and
probability of success p, the mean and standard deviation of X
are:
μ X  np
 X  np(1  p)
Note: These formulas work ONLY for binomial distributions. They
can’t be used for other distributions!
3
Binomial Formula
We can find the probability that a binomial random variable takes on any
of its values by adding probabilities for the different ways of getting
exactly that many successes in n trials – this involves counting with the
binomial coefficient.
The number of ways of arranging k successes among n trials is
given by the binomial coefficient
n 
n!

 
k  k!(n  k)!
for k = 0, 1, 2, …, n.
Note: n! = n(n – 
1)(n – 2)•…•(3)(2)(1) and 0! = 1.
4
Binomial Probability
The binomial coefficient counts the number of different ways in which
k successes can be arranged among n trials. The binomial probability
P(X = k) is this count multiplied by the probability of any one specific
arrangement of the k successes (use the “AND” rule)
Binomial Probability
If X has the binomial distribution with n trials and probability p of
success on each trial, the possible values of X are 0, 1, 2, …, n. If k
is any one of these values,
n k
P(X  k)   p (1  p) n k
k 
5
Number of
arrangements
of k successes
Probability of k
successes
Probability of
n − k failures
Example
Each child of a particular pair of parents has probability 0.25 of having blood type O.
Suppose the parents have five children.
(a) Find the probability that exactly three of the children have type O blood.
Let X = the number of the 5 children with type O blood. We can argue that X has
a binomial distribution with n = 5 and p = 0.25.
5
P(X  3)   (0.25)3 (0.75) 2 10(0.25) 3 (0.75) 2  0.08789
3
(b) Should the parents be surprised if more than three of their children
have type O blood?

6
P(X  3)  P(X  4)  P(X  5)
5 
5
4
1
  (0.25) (0.75)   (0.25) 5 (0.75) 0
4 
5
 5(0.25) 4 (0.75)1 1(0.25) 5 (0.75) 0
 0.01465  0.00098  0.01563
Since there is only a
1.5% chance that more
than three children out
of five would have type
O blood, the parents
should be surprised!
Binomial distribution in statistical sampling
Suppose a population of S’s and F’s contains a certain
proportion p of S’s. If the population is much larger than the
sample, the count X of successes in an SRS of size n has
approximately the binomial distribution B(n, p).
The n observations will be nearly independent when the
size of the population is much larger than the size of the
sample. As a rule of thumb, the binomial sampling
distribution for counts can be used when the population
is at least 20 times as large as the sample.
Binomial mean and standard deviation
0.3
0.2
a)
0.15
0.1
0.05
0
0
1
2
0.3
s = npq = np(1 - p)
3
4
5
6
7
8
9
10
8
9
10
8
9
10
Number of successes
0.25
P(X=x)
m = np
P(X=x)
The center and spread of the
binomial distribution for a count X
are defined by the mean m and
standard deviation :
0.25
We often write q as 1 – p.
b)
0.2
0.15
0.1
0.05
0
Effect of changing p when n is fixed.
a) n = 10, p = 0.25
0
1
2
3
4
5
6
7
Number of successes
0.3
b) n = 10, p = 0.5
c) n = 10, p = 0.75
For small samples, binomial distributions
are skewed when p is different from 0.5.
P(X=x)
0.25
0.2
c)
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
Number of successes
Color blindness
The frequency of color blindness (dyschromatopsia) in the
Caucasian American male population is estimated to be
about 8%. We take a random sample of size 20 from this population.
The population is definitely larger than 20 times the sample size, thus
we can approximate the sampling distribution of X = the number of the
20 who are colorblind by B(n = 20, p = 0.08).
• What is the probability that five individuals or fewer in the sample are
color blind?
Use Table C: n=20, p=.08
P(x ≤ 5) = sum of the entries for X=0,1,2,3,4 and 5 under the p=.08 column,
n=20 page
= .1887 + .3282 + .2711 + .1414 + .0523 + .0145
= .9962
• What is the probability that more than five will be color blind?
P(x > 5) = 1  P(x ≤ 5) =1  0..9962 = 0.0038
Calculations
The probabilities for a Binomial distribution can be calculated by using JMP.
In
JMP,
Highlight a column,
click the column triangle ( ), and
click formula to open the box on the
right.
Select Discrete Probability
“Binomial Probability” for
the probability of a specific number x
of successes P(X = x)
• Or “Binomial Distribution” for
the cumulative distribution function
P(X ≤ x)
•Enter values for p, n and x as shown
in the diagram.
•Click “OK.”
•P(X≤x) is illustrated.
• Choose
Color blindness
The frequency of color blindness (dyschromatopsia) in the
Caucasian American male population is estimated to be
about 8%. We take a random sample of size 20 from this population.
What is the probability that exactly five individuals in the sample are color
blind?
Use JMP: Probability  Binomial Probability (0.08, 20, 5) = .0145449122
P(x= 5) = Binomial Probability (0.08, 20, 5) = 0.0145 – or use Table C
Or use the TI-83: 2nd -> VARS -> binompdf( 20, .08, 5 ) = .0145449122
binomcdf( 20, .08, 5 ) = P( X ≤ 5 ) = .9962005132
B(n = 20, p = 0.08)
Probability distribution and histogram for the number
of color blind individuals among 20 Caucasian males.
What are the mean and standard deviation of the count
of color blind individuals in the SRS of 20 Caucasian
American males?
µ = np = 20*0.08 = 1.6
σ = sqrt(np(1  p)) = sqrt(20*0.08*0.92) = 1.213
What if we take an SRS of size 10? Of size 75?
µ = 75*0.08 = 6
σ = √(10*0.08*0.92) = 0.86
σ = √(75*0.08*0.92) = 2.35
0.5
0.2
0.4
0.15
0.3
p = .08
n = 10
0.2
0.1
P(X=x)
P(X=x)
µ = 10*0.08 = 0.8
p = .08
n = 75
0.1
0.05
0
0
0
1
2
3
4
5
Number of successes
6
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Number of successes
Sample proportions
The proportion of “successes” can be more informative than the count.
In statistical sampling the sample proportion of successes, p̂, is used to
estimate the proportion p of successes in a population.
For any SRS of size n, the sample proportion of successes is:
pˆ =

count of successes in the sample X
=
n
n
In an SRS of 50 students in an undergrad class, 10 are Hispanic:
p̂ = (10)/(50) = 0.2 (proportion of Hispanics in sample)
The 30 subjects in an SRS are asked to taste an unmarked brand of coffee and rate it “would buy” or
“would not buy.” Eighteen subjects rated the coffee “would buy.”
p̂ = (18)/(30) = 0.6 (sample proportion of “would buy”)
If the sample size is much smaller than the size of a population with
proportion p of successes, then the mean and standard deviation of
p̂ are:
m pˆ = p

s pˆ =
p(1 - p)
n
Because the mean is p, we say that the sample proportion in an SRS is an unbiased
estimator of the population proportion p.

The variability of p-hat decreases as the sample size increases. So larger samples
usually give closer estimates of the population proportion p.
Normal approximation
If n is large, and p is not too close to 0 or 1, the binomial distribution can be
approximated by the normal distribution N(m= np,  = sqrt(np(1  p))).
Practically, the Normal approximation can be used when both np ≥10 and
n(1  p) ≥10.
If X is the count of successes in the sample and p̂ = X/n, the sample proportion
of successes, their sampling distributions for large n, are:
–
X approximately N(µ = np, σ = sqrt(np(1 − p)))
–
p̂ is approximately N (µ = p, σ = sqrt(p(1 − p)/n))
Sampling distribution of the sample proportion
The sampling distribution of p̂ is never exactly normal. But as the sample size
increases, the sampling distribution of p̂ becomes approximately normal.
The normal approximation is most accurate for any fixed n when p is close to
0.5, and least accurate when p is near 0 or near 1.
Color blindness
The frequency of color blindness (dyschromatopsia) in the
Caucasian American male population is about 8%.
We take a random sample of size 125 from this population. What is the
probability that six individuals or fewer in the sample are color blind?
– Sampling distribution of the count X: B(n = 125, p = 0.08)  np = 10
P(X ≤ 6) = use JMP = 0.1198 or about 12%
– Normal approximation for the count X: N(np = 10, √np(1  p) = 3.033)
P(X ≤ 6) = use JMP: Normal Distribution(6, mean=10, sd=3.033)
– Or z = (x  µ)/σ = (6 10)/3.033 = 1.32  P(X ≤ 6) = 0.0934 from Table A or
JMP
The normal approximation is reasonable, though not perfect. Here p = 0.08 is not
at all close to 0.5 when the normal approximation is at its best.
A sample size of 125 is the smallest sample size that can allow use of the normal
approximation (to get both np = 10 and n(1  p) = 115 greater than 10).
Normal approximation: continuity correction
The normal distribution is a better approximation of the binomial
distribution, if we perform a continuity correction where x’ = x + 0.5 is
substituted for x, and P(X ≤ x) is replaced by P(X ≤ x + 0.5).
Why? A binomial random variable is a discrete variable that can only
take whole numerical values. In contrast, a normal random variable is
a continuous variable that can take any numerical value.
P(X ≤ 10) for a binomial variable is P(X ≤ 10.5) using a normal approximation.
P(X < 10) for a binomial variable excludes the outcome X = 10, so we exclude
the entire interval from 9.5 to 10.5 and calculate P(X ≤ 9.5) when using a
normal approximation.
HW: Read section 5.5; omit section 5.6
Do Exercises #5.85, 5.87- 5.90, 5.93-5.95, 5.99,
5.100, 5.102, 5.144