Download Probability Distribution for a Discrete random variable

PubH 6414 Worksheet 6b: Binomial and Poisson Distributions
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Example 1: Probability Distribution for a Discrete random variable
Number of people in a family in the US (2005 data).
Number of family members
Probability
2
0.42
3
0.228
4
0.209
5
0.092
6
0.032
7+
0.018
Use the probability distribution table above to calculate the following probabilities for the
number of people in US families in 2005. Let X represent the number of people in a family.
a.
P(X ≥ 4)
= P(X=4) + P(X=5) + P(X=6) + P(X=7 or more)
= .209 + .092 + .032 + .018
= 0.351
b.
P(3 ≤ X ≤ 5) = P(X=3) + P(X=4) + P(X=5
= .228 + .209 +.092
= .529
c.
P(X < 4)
= P(X=2) + P(X=3)
= 0.42 + 0.228
= 0.648
Example 2: Binomial Distribution
1. Number of heads in 2 coin tosses
i. What are the 4 possible outcomes from 2 coin tosses:
HH TT HT TH
ii. Probability of 0, 1, 2 heads in 2 coin tosses calculated from the possible outcomes
Number of heads
Probability
0
1/4 = 0.25
1
2/4 = 0.5
2
1/4 = 0.25
iii. Probability of 0, 1, 2 heads calculated from the binomial distribution formula
n = 2, π = 0.5
2!
2 *1
0.50 (1 − 0.5) 2 − 0 =
* 1 * 0.25 = 1 * 0.25 = 0.25
P(X = 0) =
0!(2 − 0)!
1 * 2 *1
2!
2 *1
0.51 (1 − 0.5) 2 −1 =
* 0.5 * 0.5 = 2 * 0.25 = 0.5
P(X = 1) =
1!(2 − 1)!
1 *1
2!
2 *1
0.52 (1 − 0.5) 2 − 2 =
* 0.25 * 1 = 1 * 0.25 = 0.25
P(X = 2) =
2!( 2 − 2)!
2 * 1 *1
PubH 6414 Worksheet 6b: Binomial and Poisson Distributions
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2. Probability Distribution for X~ B (10,0.08)
Number of patients
surviving 5 years
0
1
2
3
4
5
6
7
8
9
10
Probability
< 0.0001
<0.0001
0.0001
0.0008
0.0055
0.0264
0.0881
0.2013
0.3020
0.2684
0.1074
Use the probability distribution above for X ~ B(10, 0.8) to calculate the probability that
the indicated number of patients survive at least 5 years.
i.
P(X = 8)
= 0.3020
ii.
P(X > 8)
= P(X=9) + P(X=10)
= 0.2684 + 0.1074
= 0.3758
iii.
P(X < 7)
= 1 - P(X ≥ 7)
= 1 – [ P (X=7) + P(X=8) + P(X > 8)]
= 1 – [0.2013 + 0.3020 + 0.3758]
= 1 - 0.8791
= 0.1209
PubH 6414 Worksheet 6b: Binomial and Poisson Distributions
3. Probability distribution for X ~ B(8, 0.03)
Number of deaths
from heart attack
0
Probability
1
0.194
2
0.021
3
0.0013
4
< 0.001
5
<0.001
6
<0.001
7
<0.001
8
<0.001
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0.784
Use the probability distribution above for X ~ B(8, 0.03) to calculate the probability that
the indicated number of patients die from a heart attack
i.
P(X > 1)
= 1 – P(X ≤ 1)
= 1 – [P(X=0) + P(X=1)]
= 1-[0.784 + 0.194]
= 1-0.978
= 0.022
ii.
P( X < 2)
= P(X=0) + P(X=1)
= 0.784 +0.194
= 0.978
iii.
P(1 ≤ X ≤ 2)
= P(X=1) + P(X=2)
= 0.194 +0.021
= 0.215
iv.
P(all 8 patients
survive)
= P(X = 0)
= 0.784
PubH 6414 Worksheet 6b: Binomial and Poisson Distributions
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Example 3: Normal approximation to the binomial distribution
The risk of osteoporosis being diagnosed in the next year for a group of women age 70+ with
osteopenia is 0.50.
a. For a group of 100 women in this population identify the parameters of the binomial
distribution where ‘success’ = diagnosis of osteoporosis in the next year.
n = 100
π = 0.50
mean of binomial distribution = 100*0.50 = 50
b. Is the normal approximation to the binomial distribution appropriate?
nπ = 50
n(1-π) =50
Both are > 5 so normal approximation to the binomial is appropriate.
c. What are the mean and standard deviation for this normal approximation?
mean = n*π = 50
variance = n*π(1−π) = 50*0.5 = 25
std. deviation = square root of variance = 5
d. How would you calculate the probability that between 40 and 60 women in this
population will be diagnosed with osteoporosis in the next year? Draw the normal curve
approximation, the area that corresponds to this probability.
P(40-60 women diagnosed) = 0.9772499 - 0.02275013 = 0.9544997 = 0.9545
Draw a normal distribution curve with mean at 50 and SD = 5. Notice most of the curve
is contained between 35 and 65. Draw lines at 40 and 60, which is the area between 2 SD
below the mean and 2 SD above the mean.
PubH 6414 Worksheet 6b: Binomial and Poisson Distributions
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Use Rcmdr to get the probability. If you use the menu option, you must subtract the
probability below 40 from that below 60.
Rcmdr Menu: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable Values = 40, 60; mu=50; sigma=5; select Lower Tail
R Script: pnorm(60, 50, 5)-pnorm(40, 50, 5)
Example 4: Normal approximation to the Poisson distribution
At a large urban hospital, the average number of emergency department admissions on a
Saturday night is 120. On one Saturday in July, 160 patients were admitted to the emergency
department. Use the normal approximation to the Poisson distribution to determine the
probability of admitting 160 or more patients given the expected number of admissions.
a. Is the normal approximation appropriate?
Yes, the average rate / Saturday = 120 which is > 10.
b. What is the mean of the normal approximation?
120
c. What is the standard deviation of the normal approximation?
sqrt(120) = 10.95
d. Draw the normal approximation curve and the area that represents > 160 patients
admitted.
Draw a normal distribution curve with mean at 120 and SD = 10.95. Most of the area will
be between 87.2 and 152.9. Since 160 is beyond 3 SD above the mean, the probability of
admitting 160 patients given the typical admission of 120 is very small.
PubH 6414 Worksheet 6b: Binomial and Poisson Distributions
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e. Calculate this probability.
P(admitting > 160) = 0.0001296133 = 0.00013
Rcmdr Menu: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable Value = 160; mu=120; sigma=10.95; select Upper Tail
R Script: pnorm(160, 120, 10.95, lower.tail=FALSE)
1-pnorm(160, 120, 10.95)