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The discrete random variable X has the probability function. Valueof X = x 0 1 2 3 4 5 6 7 P(x) 0 k 2k 2k 3k k2 2k2 7k2 + k Evaluate p(x < 6), p(x ≥ 6) and p(0 < x < 5)

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प्रश्न

The discrete random variable X has the probability function.

Value
of X = x
0 1 2 3 4 5 6 7
P(x) 0 k 2k 2k 3k k2 2k2 7k2 + k

Evaluate p(x < 6), p(x ≥ 6) and p(0 < x < 5)

योग
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उत्तर

P(x < 6) = P(x = 0) + P(x = 1) + P(x = 2)+ P(x = 3) + p(x = 4) + P(x = 5)

= 0 + k + 2k +2k + 3k + k2

= 8k + k2 

= `8(1/10) +  (1/10)^2`

= `8/10 + 1/100`

= `(80 + 1)/100`

P(x < 6) = `81/100`

P(x ≥ 6) = P(x = 6) + p(x = 7)

= 2k2 + 7k2 + k

= 9k2 + k

= `9(1/10)^2 + 1/10`

= `9/100 + 1/10`

= `(9 + 10)/100`

= `19/100`

∴ P(x ≥ 6) = 19/100`

P(0 < x < 5) = P(x = 1) + P(x = 2) + p(x = 3) + P(x = 4)

= k + 2k + 2k + 3k

= 8k

= `8(1/10)`

∴ P(0 < x < 5) = `8/10`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Random Variable and Mathematical expectation - Exercise 6.1 [पृष्ठ १३३]

APPEARS IN

सामाचीर कलवी Business Mathematics and Statistics [English] Class 12 TN Board
अध्याय 6 Random Variable and Mathematical expectation
Exercise 6.1 | Q 6. (ii) | पृष्ठ १३३

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