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

(

_{10}C_{x})(.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

1.00001. 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

meanandstandard deviationfor 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 approximationto 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 isnormalcdf(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

meanandstandard deviationfor 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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