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A sample of 15 data is as follows: 17, 18, 17, 17, 13, 18, 5, 5, 6, 7, 8, 9, 20, 17, 3. the mode of the data is
Mode refers the most frequently appeared value (which is 17 in this case).
In a Poisson distribution if 2P(X = 1) = P(X = 2), then what is the mean?
If X is random variable, then Poisson’s distribution is given by:
The continuous distribution variable X is distributed uniformly where X ~ U(6,12). Find the mean of X?
X ~ U(6,12)
Comparing with X ~ U(a,b)
∴ a = 6 and b = 12
Mean of uniform distributed variable is given as
Consider a discrete random variable Y that takes values 2, 1, 0, 1, and 2 with probability 0.2 each. The values of the cumulative distribution function G(y) at y = 1 and y = 1 are _____ and _____ respectively.
G(1) = p(2) + p(1)
= 0.2 + 0.2 = 0.4
G(1) = p(2) + p(1) + p(0) + p(1)
= 0.2 + 0.2 + 0.2 + 0.2
= 0.8
A die is tossed thrice. Success is getting 2 or 5 on a toss. What is the variance of the number of success?
P(success) = Probability of getting 2 or 5 is 2/6 = 1/3
P(failure) = Probability of not getting 2 or 5 is 4/6 = 2/3
Probability of 0 success and 3 failure is
Probability of 1 success and 2 failure is
Probability of 2 success and 1 failure is
Probability of 3 success and 0 failure is
x is the number of success and p(x) is probability of x times success.
What is the variance of the random variable X whose probability mass function is given below?
q^{2} + 2pq + p^{2} = 1
(p+q)^{2} = 1
∴ (p+q) = 1
= 0 × q^{2} + 1 × 2pq + 2 × p^{2}
= 2p (q+p)
∴ μ = 2p
Var(x) = 2pq + 4p^{2} − 4p^{2}
Var(x) = 2pq
3, k, 2, 8, m, 3
The arithmetic mean of the list of numbers above is 4. If k and m are integers and k ≠ m, what is the median of the list?
Mean = 4
k ≠ m so k = m = 4 is out.
{k, m} = {1, 7} or {2, 6} or {3, 5}
for median of {3, k, 2, 8, m, 3}
{1, 2, 3, 3, 7, 8} or {2, 2, 3, 3, 6, 8} or {2, 3, 3, 3, 5, 8}
Median: (3 + 3) /2 = 3
If a random variable X has a Poisson distribution with variance 6, then expectation [E(X + 3)^{2}] equals _____.
Var(X) = E(X) = 6
Var(X) = E(X^{2})  (E(X))^{2}
6 = E(X^{2}) – 36
∴ E(X^{2}) = 42
[E(X + 3)^{2}]
= E(X^{2} + 6X + 9 )
= E(X^{2}) + 6E(X) + 9
= 42 + 6× 6 + 9
= 87
The probability density function of a random variable function y has the following probability function:
find p(y ≥ 5)?
Since y is a random variable:
= 0 + m+ m + 3m + 3m + m^{2} + 4m^{2 }+ 5m^{2} + m = 1
10m^{2} + 9m−1 = 0
(m+1) (10m−1) = 0
∴ m = −1or m = 1/10
since probability cannot be in negative
In a binomial distribution, the mean is 4 and variance is 3. Then the mode is
np = 4 and npq = 3
thus q = (3/4) and p = (1q) = (1/4)
mode is an integer such that np+p > x > npq
⇒ 4 +(1/4) > x >4(3/4)
⇒ (13/4)<x<(17/4)
3.25<x<4.25
⇒ x = 4
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