Sanderson M. Smith
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SAMPLING FROM A 35% "YES" POPULATION A large population consists of 35% "YES" people. A random sample of size 10 is chosen from this population. Let x = the number of YES people in the sample.
1. What is the most likely value for x?
2. What is the probability that 4 < x < 8?
3. What is the probability that x < 2?
4. What is the probability that x is at least 8?
Analysis:
The probability that a random sample of size 10 contains x YES people is
(10Cx)(.35)x(.65)(10-x)
Here is a table of probability values for x:
Value of x Probability 0 .01346
1 .07249
2 .17565
3 .25222
4 .23767
5 .15357
6 .06891
7 .02120
8 .00428
9 .00051
10 .00003
TOTAL 1.0000
1. What is the most likely value for x?
Response: The most likely value is x = 3.
2. What is the probability that 4 < x < 8?
Response: This is the probability that x = 5 or x = 6 or x = 7. This probability is 0.15357 + 0.06891 + 0.02120 = 0.24368, or about 24.4%.
3. What is the probability that x < 2?
Response: The probability that x = 0 or x = 1 is 0.01346 + 0.07249 = 0.08595, or about 8.6%.
4. What is the probability that x is at least 8?
Response: The probability that x = 8 or x = 9 or x = 10 is 0.00428 + 0.00051 + 0.00003 = 0.00482, or approximately 0.5%.
======================================================
======================================================OK, here's a slightly more sophisticated situation relating to a 35% YES population:
A large population is 35% YES. A random sample of size 120 is chosen from the population.
Let x = the number of YESSES in the sample.
Let p(hat) = the proportion of YESSES in the sample.
Question 1:
What are the possible values of the random variable x?
Response:
0,1,2,3,...,118,119,120. The random variable x has 121 possible values.
Question 2:
What are the possible values of the random variable p(hat)?
Response:
0/120, 1/120, 2/120, 3/120,...,118/120, 119/120, 120/120. The random variable p(hat) has 121 possible values.
Question 3:
Find the mean and standard deviation for the random variable x?
Response:
The mean of x is Np = 120(.35) = 42.
The standard deviation of x is SQRT[Np(1-p)] = SQRT[120(.35)(.65)] = 5.22.Question 4:
What is the probability that the number of YESSES in the sample is at least 41 and at most 44?
Response:
One could use a binomial distribution. Using the TI-83, the probability is
binomcdf(120,.35,44) - binomcdf(120,.35,40) = 0.68638 - 0.39048 = 0.2959.Since Np and N(1-p) are both greater than 10, one could use a normal approximation to the binomial distribution. Since the normal distribution is a continuous distribution, we want the probability that x is between 40.5 and 44.5. Using the TI=83, the probability is normalcdf(40.5,44.5,42,5.22) = 0.29708. Note that this is quite close to the binomial probability.
The requested probability if approximately 30%.
Question 5:
Find the mean and standard deviation for the random variable p(hat)?
Response:
The mean of p(hat) is p = .35.
The standard deviation of p(hat) is SQRT[p(1-p)/N] = SQRT[(.35)(.65)/120] = 0.04354.Question 6:
What is the probability that the sample proportion of YESSES is between 20% and 30%?
Response:
Since Np and N(1-p) are both greater than 10, one could use a normal approximation. Using the tI-83, the requested probability is
normalcdf(.195,.305,.35,.04354) = 0.15049, or about 15%.========================================================
"Chance favors prepared minds." -LOUIS PASTEUR
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