Advertisements
Advertisements
प्रश्न
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 |
Advertisements
उत्तर १
We construct the following table to calculate E (X) and V (X) :
| X = xi | pi =P [X = xi] | xi · pi | xi 2·pi = xi × xi·pi |
| -2 | 0.2 | -0.4 | 0.8 |
| -1 | 0.3 | -0.3 | 0.3 |
| 0 | 0.1 | 0 | 0 |
| 1 | 0.15 | 0.15 | 0.15 |
| 2 | 0.25 | 0.5 | 1 |
| Total | 1 | -0.05 | 2.25 |
From the table, Σxi · pi = -0.05 and Σxi2 · pi = 2.25
∴E (X) = Σxi · pi= -0.05
and V (X) = Σxi 2 ·pi - ( Σxi · pi)2
= 2.25 - (-0.05)2
= 2.25 - 0.0025 = 2.2475
Hence, E (X)= -0.05 and V (X) = 2.2475.
उत्तर २
Expected value of X 5
= E(X) = \[\sum\limits_{i=1}^{5} x_i.\text{P}_i(x_i)\]
= (–2) x (0.2) + (–1) x (0.3) + 0 x (0.1) + 1 x (0.15) + 2 x (0.25)
= – 0.4 – 0.3 + 0 + 0.15 + 0.5
= – 0.05
E(X2) = \[\sum\limits_{i=1}^{5} x_i.\text{P}_i(x_i)\]
= (–2)2 x (0.2) + (–1)2 x (0.3) + 02 x (0.1) + 12 x (0.15) + 22 x (0.25)
= 0.8 + 0.3 + 0 + 0.15 + 1
= 2.25
∴ Variance of X
= Var(X)
= E(X2) – [E(X)]2
= 2.25 – (– 0.05)2
= 2.2475.
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) | 0.4 | 0.4 | 0.2 |
State if the following is not the probability mass function of a random variable. Give reasons for your answer.
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.1 | 0.5 | 0.2 | − 0.1 | 0.2 |
State if the following is not the probability mass function of a random variable. Give reasons for your answer.
| Y | −1 | 0 | 1 |
| P(Y) | 0.6 | 0.1 | 0.2 |
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.
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.
Find k if the following function represent p.d.f. of r.v. X
f (x) = kx, for 0 < x < 2 and = 0 otherwise, Also find P `(1/ 4 < x < 3 /2)`.
Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by f(x) = `1/5`, for 0 ≤ x ≤ 5 and = 0 otherwise.
Find the probability that waiting time is between 1 and 3.
Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by f(x) = `1/5`, for 0 ≤ x ≤ 5 and = 0 otherwise.
Find the probability that the waiting time is more than 4 minutes.
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 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) =
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) = `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) =
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 :
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
Solve the following problem :
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Identify the probability distribution of X and state the formula for p. m. f. of X.
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 (X ≤ 3/ X > 0)
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(x≤1)
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(x ≥ 1.5)
Find the probability distribution of number of number of tails in three tosses of a coin
Find expected value and variance of X, the number on the uppermost face of a fair die.
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).
Find k if the following function represents the p. d. f. of a r. v. X.
f(x) = `{(kx, "for" 0 < x < 2),(0, "otherwise."):}`
Also find `"P"[1/4 < "X" < 1/2]`
X is r.v. with p.d.f. f(x) = `"k"/sqrt(x)`, 0 < x < 4 = 0 otherwise then x E(X) = _______
Fill in the blank :
E(x) is considered to be _______ of the probability distribution of x.
Solve the following problem :
Find the expected value and variance of the r. v. X if its probability distribution is as follows.
| x | 1 | 2 | 3 | ... | n |
| P(X = x) | `(1)/"n"` | `(1)/"n"` | `(1)/"n"` | ... | `(1)/"n"` |
If X denotes the number on the uppermost face of cubic die when it is tossed, then E(X) is ______
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) = ______
If 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) is ______
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 values of discrete r.v. are generally obtained by ______
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 ≤ 4) = `square + square + square + square = square`
The following function represents the p.d.f of a.r.v. X
f(x) = `{{:((kx;, "for" 0 < x < 2, "then the value of K is ")),((0;, "otherwise")):}` ______
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)
The p.m.f. of a random variable X is as follows:
P (X = 0) = 5k2, P(X = 1) = 1 – 4k, P(X = 2) = 1 – 2k and P(X = x) = 0 for any other value of X. Find k.
