Question

When a die is thrown, x denotes the number that turns up. Find e(x), e(x2) and var(x)?

Answer 1

SOLUTION

GIVEN

When a die is thrown, X denotes the number that turns up.

TO DETERMINE

E(X) , E(X²) , Var(X)

EVALUATION

Here a die is thrown

X denotes the number that turns up.

Then the spectrum consists of the value 1 , 2 , 3 , 4 , 5 , 6

Now

E(X)

$\displaystyle \sf{ = \sum xp}$

$\displaystyle \sf{ = \bigg( 1. \frac{1}{6} + 2. \frac{1}{6} + 3. \frac{1}{6} +4. \frac{1}{6} + 5. \frac{1}{6} +6. \frac{1}{6} \bigg)}$

$\displaystyle \sf{ = \frac{1}{6} (1 + 2 + 3 + 4 + 5 + 6)}$

$\displaystyle \sf{ = \frac{21}{6} }$

$\displaystyle \sf{ = \frac{7}{2} }$

Now

E(X²)

$\displaystyle \sf{ = \sum {x}^{2} p(x)}$

$\displaystyle \sf{ = \bigg( {1}^{2} . \frac{1}{6} + {2}^{2} . \frac{1}{6} + {3}^{2} . \frac{1}{6} + {4}^{2} . \frac{1}{6} + {5}^{2} . \frac{1}{6} + {6}^{2} . \frac{1}{6} \bigg)}$

$\displaystyle \sf{ = \frac{1}{6} ( {1}^{2} + {2}^{2} + {3}^{2} + {4}^{2} + {5}^{2} + {6}^{2} )}$

$\displaystyle \sf{ = \frac{1}{6} (1 + 4+ 9 + 16+ 25 + 36)}$

$\displaystyle \sf{ = \frac{91}{6} }$

Now

Var(X)

= E(X²) - (E(X))²

$\displaystyle \sf{ = \frac{91}{6} - \frac{49}{4} }$

$\displaystyle \sf{ = \frac{182 - 147}{12} }$

$\displaystyle \sf{ = \frac{35}{12} }$

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Answer 2

Given : a die is thrown, x denotes the number that turns up

To Find :  E(x) , E(x²) and Var(x)

Solution:

x            p(x)       x.P(x)         x²P(x)

1            1/6         1/6            1/6

2           1/6         2/6           4/6

3           1/6         3/6            9/6

4           1/6          4/6          16/6

5           1/6          5/6           25/6

6           1/6           6/6          36/6

                          21/6           91/6

E(x) = ∑xp(x)    =   21/6  = 3.5

E(x²)  = ∑x²p(x)  = 91/6  = 15.167

Var(x) = e(x²) - (e(x))²

Var(X)  =  ∑x²p(x) - (∑xp(x))²    

=   91/6  -   ( 21/6)²

=  91/6 - 441/36

= 105/36

=  35/12

= 2.9167

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