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The discrete random variable X has the probability function X 1 2 3 4 P(X = x) k 2k 3k 4k Show that k = 0 1 - Business Mathematics and Statistics

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प्रश्न

The discrete random variable X has the probability function

X 1 2 3 4
P(X = x) k 2k 3k 4k

Show that k = 0 1

योग
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उत्तर

P(x = 1) = k

p(x = 2) = 2k

p(x = 3) = 3k

P(x = 4) = 4k

Since P(X = x) is a probability Mass function

`sum_(x = 1)^4` P(X = x) = 1

`sum_("i" = 1)^oo` P(xi) = 1

p(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) = 1

P(x = 1) + P(x = = 2) + P(x = 3) + P(x = 4) = 1

k + 2k + 3k + 4k = 1

10k = 1

k = `1/10`

∴ k = 0.1

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Random Variable
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Random Variable and Mathematical expectation - Exercise 6.1 [पृष्ठ १३२]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 12 TN Board
अध्याय 6 Random Variable and Mathematical expectation
Exercise 6.1 | Q 4 | पृष्ठ १३२

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