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The following table is describing about the probability mass function of the random variable X x 3 4 5 P(x) 0.2 0.3 0.5 Find the standard deviation of x.

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प्रश्न

The following table is describing about the probability mass function of the random variable X

x 3 4 5
P(x) 0.2 0.3 0.5

Find the standard deviation of x.

योग
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उत्तर

E(X) = ∑xP(x)

= (3 × 0.2) + (4 × 0.3) + (5 × 05)

= 0.6 + 1.2 + 2.5

= 4.3

E(X2) = ∑x2P(x)

= (32 × 0.2) + (42 × 0.3) + (52 × 05)

= (9 × 0.2) + (16 × 0.3) + (25 × 0.5)

= 1.8 + 4.8 + 12.5

= 19.1

Variance of the Random Variable

σ2 = Var(X) = E(X2) − (E(X))2

Var(X) = 19.1 − (4.3)2

= 19.1 − 18.49

= 0.61

Standard Deviation (σ)

σ = `sqrt(Var(X))`

= `sqrt0.61`

= 0.781

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Mathematical Expectation
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Random Variable and Mathematical expectation - Exercise 6.2 [पृष्ठ १४०]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 12 TN Board
अध्याय 6 Random Variable and Mathematical expectation
Exercise 6.2 | Q 3 | पृष्ठ १४०

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