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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: - Mathematics

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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:

पर्याय

  • `1/15`

  • `1/10`

  • `1/3`

  • `2/3`

MCQ
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उत्तर

`2/3`

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Probability Distributions - Exercise 11.6 [पृष्ठ २२०]

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

संबंधित प्रश्‍न

Suppose error involved in making a certain measurement is continuous r.v. X with p.d.f.

f (x) = k `(4 – x^2 )`, for –2 ≤ x ≤ 2 and = 0 otherwise.

P(x > 0)


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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(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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Fill in the blank :

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X = x 0 1 2 3
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Then the value of k is


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x 1 2 3 4 5
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Find P(2 ≤ X < 5)


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x 1 2 3 4 5
F(x) k2 2k2 3k2 2k 3k

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then, F(1.5) - F(- 0.5) = ?


Choose the correct alternative:

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Choose the correct alternative:

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f(x) = `{(kx^2","      0 ≤ x ≤ 2), (0","         "othenwise"):}`

Then, the probability that the lecture ends within 1 minute of the bell ringing is ______


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Then P(1 < x ≤ 4) = ______ 


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