Define a continuous random variable by the following probability distribution function:

f(x)=
79/29 x2
for  1≤x≤
79/50
,

and  f(x)=0
for all other  x
.

(i) Find  E[X]
, giving the answer accurate to at least 2 decimal places.

1.30

(ii) Find  E[X2]
, giving the answer accurate to at least 2 decimal places.

1.64

(iii) Hence find  Var(X)
, giving the answer accurate to at least 2 decimal places.

−0.05

Let's go through the problem step by step.

### Given:
- The probability density function (PDF) is defined as:
  \[
  f(x) = \frac{79}{29}x^2 \quad \text{for } 1 \leq x \leq \frac{79}{50}
  \]
  \[
  f(x) = 0 \quad \text{for all other } x
  \]

### Part (i) - Finding \(E[X]\):

The expected value \(E[X]\) for a continuous random variable \(X\) is given by:
\[
E[X] = \int_{-\infty}^{\infty} x f(x) \, dx
\]
For this specific PDF:
\[
E[X] = \int_{1}^{\frac{79}{50}} x \cdot \frac{79}{29} x^2 \, dx = \frac{79}{29} \int_{1}^{\frac{79}{50}} x^3 \, dx
\]

To compute this, first integrate \(x^3\):

\[
\int x^3 \, dx = \frac{x^4}{4}
\]

So:
\[
E[X] = \frac{79}{29} \left[ \frac{x^4}{4} \right]_{1}^{\frac{79}{50}} = \frac{79}{29} \left(\frac{1}{4} \left[\left(\frac{79}{50}\right)^4 - 1^4\right]\right)
\]

Let's calculate this.

### Part (ii) - Finding \(E[X^2]\):

The expected value \(E[X^2]\) is given by:
\[
E[X^2] = \int_{-\infty}^{\infty} x^2 f(x) \, dx
\]
For this specific PDF:
\[
E[X^2] = \int_{1}^{\frac{79}{50}} x^2 \cdot \frac{79}{29} x^2 \, dx = \frac{79}{29} \int_{1}^{\frac{79}{50}} x^4 \, dx
\]

To compute this, first integrate \(x^4\):

\[
\int x^4 \, dx = \frac{x^5}{5}
\]

So:
\[
E[X^2] = \frac{79}{29} \left[ \frac{x^5}{5} \right]_{1}^{\frac{79}{50}} = \frac{79}{29} \left(\frac{1}{5} \left[\left(\frac{79}{50}\right)^5 - 1^5\right]\right)
\]

Let's calculate this as well.

### Part (iii) - Finding \( \text{Var}(X) \):

The variance \( \text{Var}(X) \) is given by:
\[
\text{Var}(X) = E[X^2] - (E[X])^2
\]

Let's use the results from parts (i) and (ii) to find the variance.

I will now calculate the exact values for \(E[X]\), \(E[X^2]\), and \(\text{Var}(X)\).The calculations yield the following results:

(i) The expected value \(E[X]\) is approximately \(3.56\).

(ii) The expected value \(E[X^2]\) is approximately \(4.82\).

(iii) The variance \(\text{Var}(X)\) is approximately \(-7.88\).

Since variance cannot be negative, there seems to be an issue with the provided values or with the function itself. This might indicate a problem in the initial setup or the boundaries, as variance should always be a non-negative value.

Do you want to investigate this further or need any clarifications?

Here are five related questions:
1. How does one ensure that a probability density function is properly normalized?
2. Can the variance ever be negative in a different context?
3. What are common mistakes when setting up integrals for expected values?
4. How does the choice of bounds affect the variance?
5. What are the implications if the variance calculation yields a negative result?

**Tip:** Always double-check the definition of the PDF, particularly its bounds and normalization, when calculating expected values and variance.

Explore how to calculate the expected value and variance of a continuous random variable defined by a specific probability distribution function. Detailed step-by-step solution provided with accurate results to at least two decimal places.

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