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Solutions are given at the end.

Sampling Distribution of a Mean

Consider again the probability distribution for a single observation for the tree volume survey from lecture (i.e., a population distribution), but with different probabilities. Here \(x\) denotes the volume of a randomly selected tree.
\(x\) \(P(x)\)
20 0.7
30 0.3

Use the five-step method to derive two sampling distributions of the mean from a sample of observations (\(\bar{x}\)): one based on a sample size of \(n\) = 2, and a second based on a sample size of \(n\) = 3. (Hint: When \(n\) = 2 there are four possible samples in the sample space, and one of these samples is 20, 30. But when \(n\) = 3 there are eight possible samples, and one of these samples is 20, 20, 30.)

Sampling Distribution of a Median

In the previous problem you derived the sampling distribution of the mean volume of a random sample of \(n\) = 3 observations of the volumes of trees. Now use the five-step method to derive the sampling distribution of the median volume of a random sample of \(n\) = 3 observations of the volumes of trees. Recall that the median is defined as the middle observation when the observations are arranged in increasing order. (Hint: You can use the same sample space that you obtained when you were deriving the sampling distribution of the mean.)

Sampling Distribution of a Proportion

  1. Consider again the population distribution from lecture for the preference of one female platy fish, but with different probabilities.

    \(x\)

    \(P(x)\)

    C

    0.4

    Y

    0.6

    Here \(C\) and \(Y\) represent a preference for the clear-tailed and the yellow-tailed male, respectively. Use the five-step method to derive the sampling distribution of the proportion (\(\hat{p}\)) of platy fish in a sample of \(n\) = 2 observations that show a preference for the yellow-tailed male. Then use the five-step method to derive the sampling distribution when the sample size is \(n\) = 3.

  2. The number of observations on which the female platy prefers the yellow-tailed male has a binomial distribution. Note that we define a “success” as a female preferring the yellow-tailed male, and so the probability of a success is as given by the population distribution. Derive the sampling distribution for the proportion of observations out of \(n\) = 2 on which the female platy prefers the yellow-tailed male using the formula for the binomial distribution. Then use the binomial distribution formula to derive the sampling distribution for a sample size of \(n\) = 3. Note that you should get the same sampling distributions using the formula for the binomial distribution as you did when you used the five-step method in the previous problem.

Sampling Distribution of a Mean (Solution)

The sampling distribution of \(\bar{x}\) when \(n\) = 2 is shown in the following table.
\(\bar{x}\) \(P(\bar{x})\)
20 0.49
25 0.42
30 0.09
The sampling distribution of \(\bar{x}\) when \(n\) = 3 is shown in the following table.
\(\bar{x}\) \(P(\bar{x})\)
20.00 0.343
23.33 0.441
26.67 0.189
30.00 0.027

Note that the sample mean has been rounded in the case when \(n\) = 3.

Sampling Distribution of a Median (Solution)

The following table shows the sampling distribution of the median. Here I am using \(m\) to represent the median.
\(m\) \(P(m)\)
20 0.784
30 0.216

Sampling Distribution of a Proportion (Solution)

The following table shows the sampling distribution of the proportion when \(n\) = 2.
\(\hat{p}\) \(P(\hat{p})\)
0.0 0.16
0.5 0.48
1.0 0.36
The following table shows the sampling distribution of the proportion when \(n\) = 3.
\(\hat{p}\) \(P(\hat{p})\)
0 0.064
1/3 0.288
2/3 0.432
1 0.216