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

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\neq 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

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)

$6.42

step-by-step explanation: