Sanderson M. Smith

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%.

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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%.

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"Chance favors prepared minds."

-LOUIS PASTEUR