Question

Ex. (7). From a survey of 20 families in a society, the following data was
obtained :
No. of children
0
1
1
2
3
4
No. of families
5
11
2
0
2
For the random variable X = number of children in a randomly
chosen family, Find E(X) and V(X).​

Answer 1

Step-by-step explanation:

the B probability distribution table has occurred when we we try to think that when any family choosing randomly from the twenty family what are the probability among those families who having zero children who are having one children who having two children who having 3 and 4

the probability of getting a family who having one children is 11 by 20 family of getting probability of a family among 20, having three children is zero and probability of a having families who have four children is 2/20 and so on.

hope you got the answer.

Answer 2

Correct Question-

From a survey of 20 families in a society, the following data was

obtained :

No. of children  0  1  2  3  4

No. of families  5  11  2  0  2

For the random variable X = number of children in a randomly

chosen family, Find E(X) and V(X).​

Answer-

E(x)= 1.15 & V(x)= 1.2775

Step-by-step explanation:

Given:

No. of children=  0  1  2  3  4

No. of families= 5  11  2  0 2

To find:

E(X) & V(X)

Solution:

The total families are 20, so N=20

To find the probablities of the given data,

Probability [P(X=x)] = frequency/N

therefore, P(x=0) = 5/20

P(x=1) = 11/20

P(x=2) = 2/20

P(x=3) = 0/20

P(x=4) = 2/20

E(x)= µ= $\sum [x P(x)]$

E(x)= $[0 (\frac{5}{20})+ 1(\frac{11}{20})+2(\frac{2}{20})+3 (\frac{0}{20})+4(\frac{2}{20})]$

E(x)= $[0+\frac{11}{20} +\frac{4}{20} +0+\frac{8}{20}]$

E(x)= $[\frac{23}{20} ]$

E(x)= $1.15$

Now, V(x)= $\sum x^{2} P(x)- \µ^{2}$

V(x)= ${[0^{2} (\frac{5}{20} )+1^{2} (\frac{11}{20} )+2^{2} (\frac{2}{20} )+3^{2} (\frac{0}{20} )+4^{2} (\frac{2}{20} )]- (1.15)^{2} }$

V(x)= $[0+\frac{11}{20} +\frac{8}{20} +0+\frac{32}{20} ]-1.3225$

V(x)= $[\frac{52}{20} -1.3225]$

V(x)= $2.6-1.3225$

V(x)= $1.2775$

Hence, E(x)= 1.15 and V(x)= 1.2775

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