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The distribution of the amount of gravel (in tons) sold by a particular construction supply company in a given week is a continuous variable X with pdf

f(x)={
3/2(1−x2) 0≤x≤1
0 otherwise

(a) Find the Cumulative distribution function. (b) Find the E[X] and Variance (c) Find the P(.2

1 Answer

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Final answer:

The cumulative distribution function (CDF) is found by integrating the probability density function (PDF) over the desired range. Expected value (E[X]) is calculated by integrating x times the PDF, and variance is calculated by subtracting the squared expected value from the expected value of the squared variable. To find P(X > 0.2), we use the complement rule.

Step-by-step explanation:

(a) Cumulative Distribution Function (CDF):

The cumulative distribution function (CDF), denoted as F(x), of a continuous random variable X is defined as the probability that X takes on a value less than or equal to x. In this case, we have:

F(x) = ∫[0,x] f(t) dt

The limits of integration are from 0 to x, and f(t) is the probability density function (PDF) of X. Since the PDF is given by f(x) = 3/2(1−x^2) for 0≤x≤1 and 0 otherwise, we can substitute this into the integral:

F(x) = ∫[0,x] 3/2(1−t^2) dt

To find the CDF, we evaluate this integral:

F(x) = ∫[0,x] 3/2(1−t^2) dt = [3/2(t−t^3/3)]|0^x = 3/2(x−x^3/3)

(b) Expected Value (E[X]) and Variance:

The expected value or mean of a continuous random variable X, denoted as E[X], can be calculated using the formula:

E[X] = ∫x f(x) dx

Again, using the PDF f(x) = 3/2(1−x^2), we substitute this into the integral:

E[X] = ∫x 3/2(1−x^2) dx = 3/2(x−x^3/3)

To find the variance of X, denoted as Var(X), we use the formula:

Var(X) = E[X^2] − (E[X])^2

We need to evaluate ∫x^2 f(x) dx to find E[X^2].

After evaluating the integral, we can calculate the variance.

(c) To find P(X > 0.2), we need to use the complement rule:

P(X > 0.2) = 1 − P(X ≤ 0.2)

We can substitute the CDF function from part (a) into this expression to find the probability.

User Abbas Jafari
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