Question

Suppose that a survey is taken of a sample of portuguese parents with at least one child under the age of 20 living at home. the parents are asked to report the number of days per week, , in which they have family dinner at home. identify which of the tables are valid probability models for . 0 1 2 3 4 5 6 7 (x) 0.100.10 0.050.05 0.100.10 0.200.20 00 0.100.10 0.200.20 0.250.25 0 1 2 3 4 5 6 7 (x) 0.1250.125 0.1250.125 0.1250.125 0.1250.125 0.1250.125 0.1250.125 0.1250.125 0.1250.125 0 1 2 3 4 5 6 7 (x) 0.100.10 0.150.15 0.200.20 0.250.25 0.300.30 0.350.35 0.400.40 0.450.45 0 1 2 3 4 5 6 7 (x) 00 1010 1010 1010 1010 1010 2525 2525 0 1 2 3 4 5 6 7 (x) −0.50−0.50 0.200.20 0.200.20 0.200.20 0.200.20 0.200.20 0.200.20 0.200.20

Answer 1

The valid probability models for the number of days per week (x) on which Portuguese parents have family dinner at home are Tables 1 and 3. These tables satisfy the conditions of a probability distribution, with each probability being between 0 and 1, and the sum of all probabilities being equal to 1.

Step-by-step explanation:

1. Table 1: Each probability in Table 1 is positive, between 0 and 1, and the sum of all probabilities equals 1:

`\(P(X = 0) = 0.10\)`

`\(P(X = 1) = 0.05\)`

`\(P(X = 2) = 0.10\)`

`\(P(X = 3) = 0.20\)`

`\(P(X = 4) = 0.10\)`

`\(P(X = 5) = 0.20\)`

`\(P(X = 6) = 0.25\)`

`\(P(X = 7) = 0.05\)`

The sum of these probabilities is
`\(0.10 + 0.05 + 0.10 + 0.20 + 0.10 + 0.20 + 0.25 + 0.05 = 1\),` making it a valid probability distribution.

2. Table 2: This table is not a valid probability model because the sum of probabilities is greater than 1. The total probability is
`\(10 + 10 + 10 + 10 + 10 + 10 + 25 + 25 = 110\%,\)` which exceeds 100%. This violates the fundamental rule that the sum of probabilities in a distribution must equal 1.

3. Table 3: Similar to Table 1, Table 3 is a valid probability model. The probabilities are non-negative, each falling between 0 and 1, and their sum is equal to 1. This ensures that the distribution accurately represents the likelihood of having family dinner a certain number of days per week.

4. Table 4: This table is also invalid as the probability values are not between 0 and 1, and the sum exceeds 1. Negative probabilities are not meaningful in the context of this survey.

In summary, Tables 1 and 3 serve as valid probability models for the number of days per week with family dinner, adhering to the principles of probability distributions. Tables 2 and 4, on the other hand, do not meet these criteria and are not valid probability models.

Answer 2

Step-by-step explanation:

For a valid probability model for X.

For X=0,1,2,3,4,5,6,7

\small 0\leq P(X)\leq 1 (as P(X) is Probability ; Probability can only between 0 and 1).

On this Count ;

these options are not valid ; 3 and 5

X 0 1 2 3 4 5 6 7

P(X) 0 l0 l0 l0 l0 l0 25 25

X 0 I 2 3 4 5 6 7

P(X) -0.50 0.20 0.20 0.20 0.20 0.20 0.20 0.20

And also for a valid probability model for X.

\small \sum_{X=0}^{7}P(X) =1

For option 1:

\small \sum_{X=0}^{7}P(X) =0.125+0.125+0.125+0.125+0.125+ 0.125+0.125+0.125=1

Option1 satisfies both the conditions

Option 2 :

\small \sum_{X=0}^{7}P(X) =0.10+0.05+0.10+0.20+0+ 0.10+0.20+0.25=1

Option 2 satisfies both the conditions

For Option 4,

\small \sum_{X=0}^{7}P(X) =0.10+0.15+0.20+0.25+0.30+ 0.35+0.40+0.45=2.2\small \sum_{X=0}^{7}P(X) =2.2\\eq 1

Hence option 4 does not satisfy condition 2

SO , the correct answer is option 1 and 2

X 0 1 2 3 4 5 6 7

PX) 0.125 0.125 0.125 O.125 0.125 0.125 0.125 0.123

X 0 1 2 3 4 5 6 7

PX) 0.10 0.05 0.10 0.20 0 0.10 0.20 0.25