Advertisements
Advertisements
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.
APPEARS IN
संबंधित प्रश्न
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 )
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.
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
Find probability that X is negative
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).
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.
The expected value of the sum of two numbers obtained when two fair dice are rolled is ______.
If r.v. X assumes values 1, 2, 3, ……. n with equal probabilities then E(X) = `("n" + 1)/(2)`
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)` |
If X denotes the number on the uppermost face of cubic die when it is tossed, then E(X) is ______
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
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 |
Complete the following activity.
Solution: Since `sum"p"_"i"` = 1
P(X ≤ 4) = `square + square + square + square = square`
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`
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")):}` ______
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) = _______.
The value of discrete r.v. is generally obtained by counting.
