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प्रश्न
X : is number obtained on upper most face when a fair die is thrown then E(X) = ______
पर्याय
3.0
3.5
4.0
4.5
उत्तर
X : is number obtained on upper most face when a fair die is thrown then E(X) = 3.5
APPEARS IN
संबंधित प्रश्न
The number of complaints which a bank manager receives per day is a Poisson random variable with parameter m = 4. Find the probability that the manager will receive -
(a) only two complaints on any given day.
(b) at most two complaints on any given day
[Use e-4 =0.0183]
If a random variable X follows Poisson distribution such that P(X = l) =P(X = 2), then find P(X ≥ 1). [Use e-2 = 0.1353]
If X has Poisson distribution with parameter m = 1, find P[X ≤ 1] [Use `e^-1 = 0.367879`]
If X has a Poisson distribution with variance 2, find P (X = 4)
[Use e-2 = 0.1353]
If X has a Poisson distribution with variance 2, find
Mean of X [Use e-2 = 0.1353]
If X has Poisson distribution with m = 1, then find P(X ≤ 1) given e−1 = 0.3678
If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e−3 = 0.0497
The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives only two complaints on a given day
A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows Poisson distribution with mean 1.5. Find the probability that (i) no car is used on a given day, (ii) some demand is refused on a given day, given e−1.5 = 0.2231.
It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has exactly 5 rats inclusive. Given e-5 = 0.0067.
It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has more than 5 rats inclusive. Given e-5 = 0.0067.
If E(X) = m and Var (X) = m then X follows ______.
Solve the following problem :
If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, find variance of X.
The expected value of the sum of two numbers obtained when two fair dice are rolled is ______.
Choose the correct alternative:
A distance random variable X is said to have the Poisson distribution with parameter m if its p.m.f. is given by P(x) = `("e"^(-"m")"m"^"x")/("x"!)` the condition for m is ______
State whether the following statement is True or False:
X is the number obtained on upper most face when a die is thrown, then E(x) = 3.5
State whether the following statement is True or False:
A discrete random variable X is said to follow the Poisson distribution with parameter m ≥ 0 if its p.m.f. is given by P(X = x) = `("e"^(-"m")"m"^"x")/"x"`, x = 0, 1, 2, .....
State whether the following statement is True or False:
If n is very large and p is very small then X follows Poisson distribution with n = mp
The probability that a bomb will hit the target is 0.8. Using the following activity, find the probability that, out of 5 bombs, exactly 2 will miss the target
Solution: Let p = probability that bomb miss the target
∴ q = `square`, p = `square`, n = 5.
X ~ B`(5, square)`, P(x) = `""^"n""C"_x"P"^x"q"^("n" - x)`
P(X = 2) = `""^5"C"_2 square = square`
If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, using the following activity find the value of m.
Solution: X : Follows Poisson distribution
∴ P(X) = `("e"^-"m" "m"^x)/(x!)`, P(X = 1) = 0.4 and P(X = 2) = 0.2
∴ P(X = 1) = `square` P(X = 2).
`("e"^-"m" "m"^x)/(1!) = square ("e"^-"m" "m"^2)/(2!)`,
`"e"^-"m" = square "e"^-"m" "m"/2`, m ≠ 0
∴ m = `square`
State whether the following statement is true or false:
lf X ∼ P(m) with P(X = 1) = P(X = 2) then m = 1.
If X – P(m) with P(X = 1) = P(X = 2), then find the mean and P(X = 2) given that e–2 = 0.1353
Solution: Since P(X = 1) = P(X = 2)
`(e^-mm^1)/(1!) = square`
∴ m = `square`
∴ mean = `square` = `square`
Then P(X = 2) = `square` = `square`
If X ∼ P(m) with P(X = 1) = P(X = 2), then find the mean and P(X = 2).
Given e–2 = 0.1353
Solution: Since P(X = 1) = P(X = 2)
∴ `("e"^square"m"^1)/(1!) = ("e"^"-m""m"^2)/square`
∴ m = `square`
∴ P(X = 2) = `("e"^-2. "m"^2)/(2!)` = `square`
