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

b

step-by-step explanation:

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.