Question

4. The following is p.m.f. of r.v.X.
2
X=x
P(X=x)
Find P(X <3).
1
0.1
3
0.5
4
0.2
0.2.
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Answer 1

Answer:

i) Since the condition of probability mass function  ∑ ∞ i = 1 ∑i=1∞ P(xi) = 1 ∑ 7 i = 0 ∑i=07 P(xi) = 1 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1  ⇒ 10k2 + 9k – 1 = 0  ⇒ (10k – 1) (k + 1) = 0  ⇒ k = 1/10 and k = -1  Since p(x) cannot be negative, k = -1 is not applicable.  Hence k = 1/10 (ii) 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  (∵ k = 1/10)  = 8/10 + 1/100 = 81  Method 2:  P(X < 6) = 1 – P(X ≥ 6)  = 1 – [P(X = 6) + P (X = 7)]  = 1 – [2k2 + 7k2 + k]  = 1 – 9k2 – k P(0 < X < 5) = P(X = 1) + P(X = 2) + P (X = 3) + P(X = 4) = k + 2k + 2k + 3k = 8k  = 8/10 (iii) P(X ≤ x) > 1/2 To find the minimum value of x, let us construct the cumulative distribution function of X. From the table we see that P(X ≤ 0) = 0  P(X ≤ 1) = 1/10 = 0.1  P(X ≤ 2) = 3/10 = 0.3  P(X ≤ 3) = 5/10 = 0.5  P(X ≤ 4) = 8/10 = 0.8  So x = 4Read more on Sarthaks.com - https://www.sarthaks.com/874615/a-random-variable-has-the-following-probability-function-value-of-5-6-7-p-x-0-k-2k-3k-4k-5k-6k-7k