Question

Consider the population of four juvenile condors. Their weights in pounds are : 4, 5, 7, 12 (a) Let x be the weight of a juvenile condor. Write the possible unique values for x: (NOTE: Separate each value in the list with a comma.) . (b) Find the mean of the population: (c) Let x¯ be the average weight from a sample of two juvenile condors. List all possible outcomes for x¯. (If a value occurs twice, make sure to list it twice.) This is the sampling distribution for samples of size 2: (NOTE: Separate each value in the list with a comma.) . (d) Find the mean of the sampling distribution: Note: You can earn partial credit on this problem.

Answer 1

Answer:

a) 4, 5, 7, 12

b) 7

c) 4.5, 6.5, 8, 6, 8.5, 9.5

d) 7.167              

Step-by-step explanation:

We are given the following in the question:

4, 5, 7, 12

a) unique values for x

4, 5, 7, 12

b) mean of the population

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$\mu =\displaystyle\frac{28}{4} = 7$

c) sampling distribution for samples of size 2

Sample size, n = 2

Possible samples of size 2 are (4,5),(4,7),(4,12),(5,7),(5,12),(7,12)

Sample means are:

$\bar{x_1} = \dfrac{4+5}{2} = 4.5\\\\\bar{x_2} = \dfrac{4+7}{2} = 6.5\\\\\bar{x_3} = \dfrac{4+12}{2} = 8\\\\\bar{x_4} = \dfrac{5+7}{2} = 6\\\\\bar{x_5} = \dfrac{5+12}{2} = 8.5\\\\\bar{x_6} = \dfrac{7+12}{2} = 9.5$

Thus, the list is 4.5, 6.5, 8, 6, 8.5, 9.5

d) mean of the sampling distribution

$\bar{x} = \dfrac{4.5 + 6.5 + 8+ 6 + 8.5+ 9.5}{6} = \dfrac{43}{6} = 7.167$