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Choose the correct alternative: The probability mass function of a random variable is defined as: x – 2 – 1 0 1 2 f(x) k 2k 3k 4k 5k Then E(X ) is equal to:

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प्रश्न

Choose the correct alternative:

The probability mass function of a random variable is defined as:

x – 2 – 1 0 1 2
f(x) k 2k 3k 4k 5k

Then E(X ) is equal to:

विकल्प

  • `1/15`

  • `1/10`

  • `1/3`

  • `2/3`

MCQ
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उत्तर

`2/3`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Probability Distributions - Exercise 11.6 [पृष्ठ २२०]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 11 Probability Distributions
Exercise 11.6 | Q 17 | पृष्ठ २२०

संबंधित प्रश्न

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P(x > 0)


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f (x) = `x/8`, for 0 < x < 4 and = 0 otherwise.

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Determine c.d.f. of X hence find

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Determine c.d.f. of X hence find P( x < –2)


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Determine c.d.f. of X hence find P(1 < x < 2)


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f (x) = `x^2/ 3` , for –1 < x < 2 and = 0 otherwise

Determine c.d.f. of X hence find P( X > 0)


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= 0, otherwise

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