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Tamil Nadu Board of Secondary EducationHSC Commerce Class 12

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 If P(X ≤ x) > 12, then find the minimum value of x.

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Question

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

If P(X ≤ x) > `1/2`, then find the minimum value of x.

Sum
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Solution

We want the minimum value of x for which P(X ≥ x) > `1/2`

Now P(X ≤ 0) = `0 < 1/2`

P(X ≤ 1) = P(x = 0) + P(X = 1) = 0 + k

= `1/10 < 1/2`

P( X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= 0 + k + 2k = 3k

= `3/10 = 1/2`

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3)

= 0 + k + 2k + 2k = 5k

= `5/10 = 1/2`

P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3) + P(X = 4)

= 0 + k + 2k + 2k + 3k = 8k

= `8/10 > 1/2`

This Shows that the minimum value of X for which P(X ≤ x) > `1/2` is 4

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Random Variable
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Chapter 6: Random Variable and Mathematical expectation - Exercise 6.1 [Page 133]

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Samacheer Kalvi Business Mathematics and Statistics [English] Class 12 TN Board
Chapter 6 Random Variable and Mathematical expectation
Exercise 6.1 | Q 6. (iii) | Page 133

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