Question

Consider a hypothetical population of dogs in which there are four possible weights, all of which are equally likely: 42, 48, 52, or 58 pounds. If a sample of size n =2 is drawn from this population, what is the sampling distribution of the total weight of the two dogs selected? That is, what are the possible values for the total and what are the probabilities associated with each of those values?

Answer 1

Answer:

Total ____ mean ____ total probability

84 _____ 48 ______ 1/16

90 _____ 45 ______ 1/8

94 _____ 47 ______ 1/8

96 _____48 ______ 1/16

100 ____ 50 ______ 1/4

104 ____ 52 _____ 1/ 16

106 ____ 53 ______ 1/8

110 ____55 _______ 1/8

116 ____ 58 ______ 1/16

Step-by-step explanation:

Given the weights :

42 ; 48 ; 52 ; 58

Sample size, n = 2

Number of weights = 4 ; size of sample space = 4^n = 4^2 = 16

Since the weights are distinct :

Probability of each (two Selections = 1/4 * 1/4) = 1/16

Possible selections :

(42,42),(42,48),(48,42),(42,52),(52,42),(42,58),(58,42),(48,48),(48,52),(52,48),(48,58),(58,48),(52,52),(52,58),(58,52),(58,58)

All have probabilities of 1/16

Sample total ___ sample. Mean ___ probability

84___ 42 ___ 1/16

90 ____45 _ 1/16

90 ___ 45 __1/16

94 __47 ___ 1/16

94 __ 47 __ 1/16

100 __ 50 __ 1/16

100 __ 50 __ 1/16

96 __ 48 __ 1/16

100 __ 50__1/16

100 __ 50 __ 1/16

106 __ 53 __ 1/16

106 __ 53 __ 1/16

104 __ 52 ___ 1/16

110 __ 55 ___ 1/16

110 __55 ____ 1/16

116 __ 58 ___ 1/16

Sampling distribution of total weight :

Total ____ mean ____ total probability

84 _____ 48 ______ 1/16

90 _____ 45 ______ 1/8

94 _____ 47 ______ 1/8

96 _____48 ______ 1/16

100 ____ 50 ______ 1/4

104 ____ 52 _____ 1/ 16

106 ____ 53 ______ 1/8

110 ____55 _______ 1/8

116 ____ 58 ______ 1/16

Answer 2

Using probability concepts, it is found that the distribution of total weights is given by:

$P(X = 84) = \frac{1}{16}$

$P(X = 90) = \frac{1}{8}$

$P(X = 94) = \frac{1}{8}$

$P(X = 96) = \frac{1}{16}$

$P(X = 100) = \frac{1}{4}$

$P(X = 104) = \frac{1}{16}$

$P(X = 106) = \frac{1}{8}$

$P(X = 110) = \frac{1}{8}$

$P(X = 116) = \frac{1}{16}$

• A probability is the number of desired outcomes divided by the number of total outcomes.

Considering that all weights are equally as likely, we can put it all into a "table", with $4^2 = 16$ total options.

Dog 1 - Dog 2 - Total

42 - 42 - 84

42 - 48 - 90

42 - 52 - 94

42 - 58 - 100

48 - 42 - 90

48 - 48 - 96

48 - 52 - 100

48 - 58 - 106

52 - 42  - 94

52 - 48 - 100

52 - 52 - 104

52 - 58 - 110

58 - 42 - 100

58 - 48 - 106

58 - 52 - 110

58 - 58 - 116

Thus, the possible outcomes are:

• 1 for 84, 96, 104, 116.
• 2 for 90, 94, 106, 110.
• 4 for 100.

Thus, the distribution is:

$P(X = 84) = \frac{1}{16}$

$P(X = 90) = \frac{1}{8}$

$P(X = 94) = \frac{1}{8}$

$P(X = 96) = \frac{1}{16}$

$P(X = 100) = \frac{1}{4}$

$P(X = 104) = \frac{1}{16}$

$P(X = 106) = \frac{1}{8}$

$P(X = 110) = \frac{1}{8}$

$P(X = 116) = \frac{1}{16}$

A similar problem is given at https://brainly.com/question/16967884