Advertisements
Advertisements
प्रश्न
Solve the following problem :
Find the expected value and variance of the r. v. X if its probability distribution is as follows.
x | – 1 | 0 | 1 |
P(X = x) | `(1)/(5)` | `(2)/(5)` | `(2)/(5)` |
उत्तर
E(X) = \[\sum\limits_{i=1}^{3} x_i\cdot\text{P}(x_i)\]
= `-1 (1/5) + 0(2/5) + 1(2/5)`
= `(1)/(5)`
E(X2) = \[\sum\limits_{i=1}^{3} x_i^2\text{P}(x_i)\]
= `(-1)^2 (1/5) + 0^2(2/5) + 1^2(2/5)`
= `(3)/(5)`
∴ Var(X) = E(X2) – [E(X)]2
= `(3)/(5) - (1/5)^2`
= `(14)/(25)`..
APPEARS IN
संबंधित प्रश्न
State if the following is not the probability mass function of a random variable. Give reasons for your answer.
X | 0 | 1 | 2 |
P(X) | 0.1 | 0.6 | 0.3 |
State if the following is not the probability mass function of a random variable. Give reasons for your answer
Z | 3 | 2 | 1 | 0 | −1 |
P(Z) | 0.3 | 0.2 | 0.4 | 0 | 0.05 |
State if the following is not the probability mass function of a random variable. Give reasons for your answer.
X | 0 | -1 | -2 |
P(X) | 0.3 | 0.4 | 0.3 |
A random variable X has the following probability distribution:
X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
P(X) | 0 | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
Determine:
- k
- P(X < 3)
- P( X > 4)
Find the mean number of heads in three tosses of a fair coin.
The following is the p.d.f. of r.v. X :
f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise
P ( 1 < x < 2 )
If a r.v. X has p.d.f.,
f (x) = `c /x` , for 1 < x < 3, c > 0, Find c, E(X) and Var (X).
Choose the correct option from the given alternative:
If a d.r.v. X takes values 0, 1, 2, 3, . . . which probability P (X = x) = k (x + 1)·5 −x , where k is a constant, then P (X = 0) =
Choose the correct option from the given alternative:
If p.m.f. of a d.r.v. X is P (X = x) = `x^2 /(n (n + 1))`, for x = 1, 2, 3, . . ., n and = 0, otherwise then E (X ) =
Solve the following :
Identify the random variable as either discrete or continuous in each of the following. Write down the range of it.
The person on the high protein diet is interested gain of weight in a week.
70% of the members favour and 30% oppose a proposal in a meeting. The random variable X takes the value 0 if a member opposes the proposal and the value 1 if a member is in favour. Find E(X) and Var(X).
Given that X ~ B(n,p), if n = 25, E(X) = 10, find p and Var (X).
If F(x) is distribution function of discrete r.v.x with p.m.f. P(x) = `(x - 1)/(3)` for x = 0, 1 2, 3, and P(x) = 0 otherwise then F(4) = _______.
State whether the following is True or False :
If P(X = x) = `"k"[(4),(x)]` for x = 0, 1, 2, 3, 4 , then F(5) = `(1)/(4)` when F(x) is c.d.f.
Solve the following problem :
Find the expected value and variance of the r. v. X if its probability distribution is as follows.
X | 0 | 1 | 2 | 3 | 4 | 5 |
P(X = x) | `(1)/(32)` | `(5)/(32)` | `(10)/(32)` | `(10)/(32)` | `(5)/(32)` | `(1)/(32)` |
Solve the following problem :
Let X∼B(n,p) If E(X) = 5 and Var(X) = 2.5, find n and p.
If the p.m.f. of a d.r.v. X is P(X = x) = `{{:(("c")/x^3",", "for" x = 1"," 2"," 3","),(0",", "otherwise"):}` then E(X) = ______
The probability distribution of a discrete r.v.X is as follows.
x | 1 | 2 | 3 | 4 | 5 | 6 |
P(X = x) | k | 2k | 3k | 4k | 5k | 6k |
Complete the following activity.
Solution: Since `sum"p"_"i"` = 1
k = `square`
The probability distribution of a discrete r.v. X is as follows:
x | 1 | 2 | 3 | 4 | 5 | 6 |
P(X = x) | k | 2k | 3k | 4k | 5k | 6k |
- Determine the value of k.
- Find P(X ≤ 4)
- P(2 < X < 4)
- P(X ≥ 3)
The probability distribution of X is as follows:
x | 0 | 1 | 2 | 3 | 4 |
P[X = x] | 0.1 | k | 2k | 2k | k |
Find
- k
- P[X < 2]
- P[X ≥ 3]
- P[1 ≤ X < 4]
- P(2)