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प्रश्न
Fill in the blank :
E(x) is considered to be _______ of the probability distribution of x.
उत्तर
E(x) is considered to be Centre of gravity of the probability distribution of x.
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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 |
Find expected value and variance of X for the following p.m.f.
| x | -2 | -1 | 0 | 1 | 2 |
| P(X) | 0.2 | 0.3 | 0.1 | 0.15 | 0.25 |
It is known that error in measurement of reaction temperature (in 0° c) in a certain experiment is continuous r.v. given by
f (x) = `x^2 /3` , for –1 < x < 2 and = 0 otherwise
Verify whether f (x) is p.d.f. of r.v. X.
Choose the correct option from the given alternative:
If the p.d.f of a.c.r.v. X is f (x) = 3 (1 − 2x2 ), for 0 < x < 1 and = 0, otherwise (elsewhere) then the c.d.f of X is F(x) =
Let X be amount of time for which a book is taken out of library by randomly selected student and suppose X has p.d.f
f (x) = 0.5x, for 0 ≤ x ≤ 2 and = 0 otherwise.
Calculate: P(0.5 ≤ x ≤ 1.5)
X is r.v. with p.d.f. f(x) = `"k"/sqrt(x)`, 0 < x < 4 = 0 otherwise then x E(X) = _______
If F(x) is distribution function of discrete r.v.X with p.m.f. P(x) = `k^4C_x` for x = 0, 1, 2, 3, 4 and P(x) = 0 otherwise then F(–1) = _______
Solve the following problem :
The p.m.f. of a r.v.X is given by
`P(X = x) = {(((5),(x)) 1/2^5", ", x = 0", "1", "2", "3", "4", "5.),(0,"otherwise"):}`
Show that P(X ≤ 2) = P(X ≤ 3).
Solve the following problem :
The following is the c.d.f of a r.v.X.
| x | – 3 | – 2 | – 1 | 0 | 1 | 2 | 3 | 4 |
| F (x) | 0.1 | 0.3 | 0.5 | 0.65 | 0.75 | 0.85 | 0.9 | 1 |
Find the probability distribution of X and P(–1 ≤ X ≤ 2).
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)` |
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 n = 10 and E(X)= 5, find p and Var(X).
If a d.r.v. X takes values 0, 1, 2, 3, … with probability P(X = x) = k(x + 1) × 5–x, where k is a constant, then P(X = 0) = ______
Find mean for the following probability distribution.
| X | 0 | 1 | 2 | 3 |
| P(X = x) | `1/6` | `1/3` | `1/3` | `1/6` |
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 and P[X < 2]
Choose the correct alternative:
f(x) is c.d.f. of discete r.v. X whose distribution is
| xi | – 2 | – 1 | 0 | 1 | 2 |
| pi | 0.2 | 0.3 | 0.15 | 0.25 | 0.1 |
then F(– 3) = ______
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
P(X ≥ 3) = `square - square - square = 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)
