Question -
Answer -
Given a randomvariable X with its probability distribution.
(i) As we know the sumof all the probabilities in a probability distribution of a random variablemust be one.
Hence the sum ofprobabilities of given table:
⇒ 0 + k + 2k + 2k+ 3k + k2 + 2k2 + 7K2 + k = 1
⇒ 10K2 +9k = 1
⇒ 10K2 +9k – 1 = 0
⇒ (10K-1) (k + 1)= 0
k = -1, 1/10
It is known thatprobability of any observation must always be positive that it can’t benegative.
So k = 1/10
(ii) Now we have tofind P(X < 3)
P(X < 3) = P(X = 0)+ P(X = 1) + P(X = 2)
= 0 + k + 2k
= 3k
P (X < 3) = 3 ×1/10 = 3/10
(iii) Now we have findP(X > 6)
P(X > 6) = P(X = 7)
= 7K2 +k
= 7 × (1/10)2 +1/10
= 7/100 + 1/10
P (X > 6) = 17/100
(iv) Consider P (0< X < 3)
P (0 < X < 3) =P(X = 1) + P(X = 2)
= k + 2k
= 3k
P (0 < X < 3) =3 × 1/10 = 3/10