Question

(1 point) Consider the population of four juvenile condors. Their weights in pounds are : 5, 6, 10, 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.) 5, 6, 10, 12 . (b) Find the mean of the population: 8.25 (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.) 11, 15, 17, 16, 18, 22 . (d) Find the mean of the sampling distribution: 16.5

Answer 1

A

Explanation:

i took the test

Answer 2

Explanation:

A) Unique values of x = 5,6,10,12

(B) Mean of population = (sum of weights)/(total number of weights)

= (5+6+10+12)/4

= 33/4

= 8.25

(C) Possible samples of size 2 are (5,6),(6,10),(10,12),(12,5),(5,10),(6,12)

we know, mean = (sum of data values)/(total number of data values)

so,

mean 1 = (5+6)/2=11/2=5.5

mean 2 = (6+10)/2=8

mean 3 =(10+12)/2=11

mean 4 =(12+5)/2=8.5

mean 5 =(5+10)/2=7.5

mean 6 =(6+12)/2=9

So, required list is 5.5, 7.5, 8, 8.5, 9, 11

(D) mean of sample means = (sum of all sample means)/(number of sample means)

=(5.5 + 7.5 + 8 + 8.5 + 9 + 11 )/6

= 49.5/6

= 8.25