Suppose that fifteen observations are chosen at random from the pdf f_Y(y) = 3y^2, 0<=y<=1. Let X denote the number that lie in the interval (1/2, 1). Find E(X).
E(X) is the expected value or average.
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To find E(X), we need to calculate the probability of each value of X and then multiply it by the corresponding value of X.
Let's break down the problem:
- We have 15 observations chosen at random from the given probability density function (pdf), f_Y(y) = 3y^2, where 0 <= y <= 1.
- We want to find the expected value of X, which represents the number of observations that lie in the interval (1/2, 1).
To calculate E(X), we need to find the probability that any given observation lies in the interval (1/2, 1).
Let's calculate the probability of a single observation falling in this interval:
P(1/2 < Y <= 1) = ∫[1/2, 1] f_Y(y) dy
Given that f_Y(y) = 3y^2, we can evaluate this integral:
P(1/2 < Y <= 1) = ∫[1/2, 1] 3y^2 dy
= [y^3] evaluated from 1/2 to 1
= (1)^3 - (1/2)^3
= 1 - 1/8
= 7/8
Therefore, the probability of a single observation falling in the interval (1/2, 1) is 7/8.
Now, to find E(X), we multiply the probability by the value of X:
E(X) = (7/8) * 15
= 7 * 15 / 8
= 105 / 8
= 13.125
Hence, the expected value of X is 13.125.
To find the expected value of X, you need to calculate the probability of X taking on different values and then multiply each value by its corresponding probability.
In this case, X represents the number of observations that lie in the interval (1/2, 1), so X can only take on integer values from 0 to 15.
To find the probability of X = k, where k is an integer from 0 to 15, you can use the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
where n is the total number of observations (15 in this case), p is the probability that a single observation lies in the interval (1/2, 1), and C(n, k) is the binomial coefficient, which calculates the number of ways to choose k observations out of n.
In this case, p is the integral of the probability density function (pdf) f_Y(y) over the interval (1/2, 1):
p = ∫[1/2 to 1] f_Y(y) dy
= ∫[1/2 to 1] 3y^2 dy
Evaluating this integral gives:
p = [y^3] [1/2 to 1]
= (1^3 - (1/2)^3)
= 1 - 1/8
= 7/8
Now, you can calculate the expected value E(X):
E(X) = ∑[k = 0 to 15] k * P(X = k)
= ∑[k = 0 to 15] k * C(15, k) * (7/8)^k * (1/8)^(15 - k)
To calculate this sum, you can use statistical software, a calculator, or refer to binomial tables. The result will be the expected value E(X) for the given distribution.