1. Assume that x is a binomial variable with an expected value of 3 and variance of 2.4 .find (a) P(x=0) (b) P(x=12)
THnxs
To find the probabilities, we need to use the probability mass function (PMF) of a binomial distribution. The PMF is defined as:
P(x=k) = nCk * p^k * (1-p)^(n-k)
Where:
- n is the number of trials
- k is the number of successful outcomes (in this case, the value of x)
- nCk is the binomial coefficient, which represents the number of combinations of n items taken k at a time
- p is the probability of success in a single trial
In this case, we are given the expected value (mean) and variance, which can be used to determine the values of n and p. The expected value of a binomial distribution is given by E(x) = n * p, and the variance is given by Var(x) = n * p * (1-p).
Given that E(x) = 3 and Var(x) = 2.4, we can solve for n and p as follows:
E(x) = n * p
3 = n * p -- (Equation 1)
Var(x) = n * p * (1-p)
2.4 = n * p * (1-p) -- (Equation 2)
We have two equations and two unknowns (n and p), so we can solve this system of equations.
First, let's solve Equation 1 for p:
p = 3 / n -- (Equation 3)
Now, substitute Equation 3 into Equation 2:
2.4 = n * (3 / n) * (1 - (3 / n))
2.4 = 3 * (1 - (3 / n))
2.4 = 3 - (9 / n)
(9 / n) = 3 - 2.4
(9 / n) = 0.6
9 = 0.6n
n = 9 / 0.6
n = 15
Now, substitute the value of n into Equation 3 to get p:
p = 3 / n
p = 3 / 15
p = 0.2
We have now determined that the values of n and p are 15 and 0.2, respectively.
Now, let's calculate the probabilities:
(a) P(x=0):
To find P(x=0), substitute n = 15, p = 0.2, and k = 0 into the PMF formula:
P(x=0) = 15C0 * 0.2^0 * (1-0.2)^(15-0)
P(x=0) = 1 * 1 * 0.8^15
P(x=0) = 0.035
Therefore, P(x=0) is 0.035.
(b) P(x=12):
To find P(x=12), substitute n = 15, p = 0.2, and k = 12 into the PMF formula:
P(x=12) = 15C12 * 0.2^12 * (1-0.2)^(15-12)
P(x=12) = 455 * 0.2^12 * 0.8^3
P(x=12) = 0.027
Therefore, P(x=12) is 0.027.