A random variable X has the following probability distribution:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| P(X) | z | 2z | 2z | 3z | z2 | 2z2 | 7z2 + z |
Which of the following are TRUE ?
(A) P(X < 3) = 0.8
(B) P(X > 4) = 0.2
(C) The value of z is 0.1
(D) P(3 ≤ X < 5) = 0.31
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C) and (D) only
The probabilities must add up to 1:
\( \displaystyle z+2z+2z+3z+z^2+2z^2+7z^2+z=1 \)
\( \displaystyle 10z^2+9z-1=0 \)
\( \displaystyle (10z-1)(z+1)=0 \)
A probability cannot be negative, so z = 0.1. Then
\( \displaystyle P(X\lt 3)=z+2z+2z=5z=0.5 \)
\( \displaystyle P(X\gt 4)=2z^2+(7z^2+z)=0.02+0.17=0.19 \)
\( \displaystyle P(3\le X\lt 5)=3z+z^2=0.3+0.01=0.31 \)
So z = 0.1, P(X < 3) = 0.5, P(X > 4) = 0.19 and P(3 ≤ X < 5) = 0.31.
Checking the statements
(A) P(X < 3) = 0.8 — Incorrect. P(X < 3) = 5z = 0.5.
(B) P(X > 4) = 0.2 — Incorrect. P(X = 5) + P(X = 6) = 0.02 + 0.17 = 0.19.
(C) The value of z is 0.1 — Correct. \( 10z^2+9z-1=0 \) gives z = 0.1 (z = −1 is rejected).
(D) P(3 ≤ X < 5) = 0.31 — Correct. P(X = 3) + P(X = 4) = 0.3 + 0.01 = 0.31.