Advertisements
Advertisements
Question
Solve the following problem :
Let X∼B(n,p) If E(X) = 5 and Var(X) = 2.5, find n and p.
Solution
Let X ~ B(n, p)
E(X) = 5 and Var (X) = 2.5 …[Given]
But E(X) = np = 5 and
Var (X) = npq = 2.5
∴ 5(q) = 2.5
∴ q = `(2.5)/(5) = (1)/(2)`
∴ p = 1 – q = `1 - (1)/(2) = (1)/(2)`
Now, np = 5
∴ `"n"(1/2)` = 5
∴ n = 10.
APPEARS IN
RELATED QUESTIONS
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 |
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the standard deviation of X.
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
Find probability that X is negative
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 ) =
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]`
F(x) is c.d.f. of discrete 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) = _______ .
X is r.v. with p.d.f. f(x) = `"k"/sqrt(x)`, 0 < x < 4 = 0 otherwise then x E(X) = _______
Choose the correct alternative :
If X ∼ B`(20, 1/10)` then E(X) = _______
The p.m.f. of a d.r.v. X is P(X = x) = `{{:(((5),(x))/2^5",", "for" x = 0"," 1"," 2"," 3"," 4"," 5),(0",", "otherwise"):}` If a = P(X ≤ 2) and b = P(X ≥ 3), then
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 ______
If a d.r.v. X has the following probability distribution:
| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X = x) | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
then k = ______
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`
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) = _______.
