Question

assume a population of 1, 2, and 12. assume that samples of size ( n = 2 ) are randomly selected with replacement from the population. listed below are the nine different samples. complete parts a through d below.
1,1 1,2 1,12 2,1 2,2 2,12 12,1 12,2 12,12
(round to three decimal places as needed.)
b. find the standard deviation of each of the nine samples, then summarize the sampling distribution of the standard deviations in the format of a table representing the probability distribution of the distinct standard deviation values. use ascending order of the sample standard deviations.

s	probability
0.707	( \frac{2}{9} )
7.071	( \frac{2}{9} )
7.778	( \frac{2}{9} )

(type integers or fractions.)
c. find the mean of the sampling distribution of the sample standard deviations.
the mean of the sampling distribution of the sample standard deviations is ( square ).
(round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for the mean of a probability distribution

The formula for the mean \(\mu\) of a probability distribution is \(\mu=\sum(x\times P(x))\), where \(x\) are the values and \(P(x)\) are their corresponding probabilities.

Step2: Substitute the values into the formula

We have \(x_1 = 0\), \(P(x_1)=\frac{1}{3}\); \(x_2=0.707\), \(P(x_2)=\frac{2}{9}\); \(x_3 = 7.071\), \(P(x_3)=\frac{2}{9}\); \(x_4=7.778\), \(P(x_4)=\frac{2}{9}\).

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First, calculate \(0.707 + 7.071+7.778=15.556\)
Then \(\frac{15.556\times2}{9}=\frac{31.112}{9}\approx3.457\)

Answer:

\(3.457\)