Mathematics, 18.01.2021 14:00 heidibode

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?

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Answer from: phillipsangela014

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 link

Answer from: kenna162

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

96 __ 48 __ 1/16

100 __ 50__1/16

100 __ 50 __ 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 from: Quest

i believe the answer is the polygon, heptagon.

Answer from: Quest

Equatorial triangle and obtuce angle

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