Question
Assuming the true population mean $(\mu)$ is 75.47 and the population standard deviation $(\sigma)$ is 2.01 for the question 1 , calculate the following:a. How many different samples of $\mathrm{n}=3$ can be selected from the above population of 30 data points?b. What would be the grand mean for all the possible samples of $\mathrm{n}=3$ ?c. What would be the standard deviation for all the possible samples of $\mathrm{n}=3$ ?
Step 1
Thus, we calculate: \[ C(30, 3) = \frac{30!}{3!(30-3)!} = \frac{30!}{3! \cdot 27!} = \frac{30 \times 29 \times 28}{3 \times 2 \times 1} = 4060 \] So, there are 4060 different samples of size \( n = 3 \). Show more…
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