Advertisements
Advertisements
Question
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.
Options
True
False
Solution
False
F (5) = 1.
APPEARS IN
RELATED QUESTIONS
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 |
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 )
The following is the p.d.f. of r.v. X:
f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise.
P(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 p.m.f. of a d.r.v. X is P (x) = `c/ x^3` , for x = 1, 2, 3 and = 0, otherwise (elsewhere) then E (X ) =
Choose the correct option from the given alternative:
If the a d.r.v. X has the following probability distribution :
| x | -2 | -1 | 0 | 1 | 2 | 3 |
| p(X=x) | 0.1 | k | 0.2 | 2k | 0.3 | k |
then P (X = −1) =
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.
Solve the following :
The following probability distribution of r.v. X
| X=x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| P(X=x) | 0.05 | 0.1 | 0.15 | 0.20 | 0.25 | 0.15 | 0.1 |
Find the probability that
X is positive
The following is the c.d.f. of 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 p.m.f. of X.
i. P(–1 ≤ X ≤ 2)
ii. P(X ≤ 3 / X > 0).
The following is the c.d.f. of 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 |
P (–1 ≤ X ≤ 2)
Find the probability distribution of number of number of tails in three tosses of a coin
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 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).
If the p.m.f. of a d.r.v. X is P(X = x) = `{{:(x/("n"("n" + 1))",", "for" x = 1"," 2"," 3"," .... "," "n"),(0",", "otherwise"):}`, then E(X) = ______
Find the expected value and variance of r.v. X whose p.m.f. is given below.
| X | 1 | 2 | 3 |
| P(X = x) | `1/5` | `2/5` | `2/5` |
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) = ______
Using the following activity, find the expected value and variance of the r.v.X if its probability distribution is as follows.
| x | 1 | 2 | 3 |
| P(X = x) | `1/5` | `2/5` | `2/5` |
Solution: µ = E(X) = `sum_("i" = 1)^3 x_"i""p"_"i"`
E(X) = `square + square + square = square`
Var(X) = `"E"("X"^2) - {"E"("X")}^2`
= `sum"X"_"i"^2"P"_"i" - [sum"X"_"i""P"_"i"]^2`
= `square - square`
= `square`
