Let X be a random variable, and let P(X = x) denote the probability that X takes the value x. Suppose that the points (x, P(X = x)), x = 0, 1, 2, 3, 4, lie on a fixed straight line in the xy-plane, and P(X = x) = 0 for all x ∈ ℝ – {0, 1, 2, 3, 4}. If the mean of X is 5/2 , and the variance of X is α, then the value of 24α is ______.

The correct answer is 42.

Step-by-step Explanation:

Let $X$ be a discrete random variable that takes values in the set $\{0,1,2,3,4\}$ with probabilities $P(X=x)$ for $x=0,1,2,3,4$.

Given that the points $(x,P(X=x\left)\right)$ lie on a fixed straight line, we can write the probability mass function of $X$ as:
$P(X=x)=mx+c$
where $m$ is the slope and $c$ is the y-intercept of the line.

Since $P(X=x)$ is a probability distribution, the sum of all probabilities must equal 1:
$\sum _{x=0}^{4}P(X=x)=1$
Substituting $P(X=x)=mx+c$:
$\sum _{x=0}^4(mx+c)=1$
$m(0+1+2+3+4)+5c=1$
$10m+5c=1$
Dividing by 5 gives our first equation:
$2m+c=\frac{1}{5}$     --- (Equation 1)

The mean of $X$ is given as $\frac{5}{2}$:
$E\left[X\right]=\sum _{x=0}^{4}x·P(X=x)=\frac{5}{2}$
Substituting $P(X=x)=mx+c$:
$\sum _{x=0}^{4}x(mx+c)=\frac{5}{2}$
$m\sum _{x=0}^4{x}^2+c\sum _{x=0}^{4}x=\frac{5}{2}$
Calculating the sum of squares:
$\sum _{x=0}^4{x}^2=0+1+4+9+16=30$
Substituting the sums:
$30m+10c=\frac{5}{2}$
Dividing by 10 gives our second equation:
$3m+c=\frac{1}{4}$     --- (Equation 2)

Subtracting Equation 1 from Equation 2:
$(3m+c)-(2m+c)=\frac{1}{4}-\frac{1}{5}$
$m=\frac{1}{20}$

Substitute $m=\frac{1}{20}$ back into Equation 1:
$2\left(\frac{1}{20}\right)+c=\frac{1}{5}$
$\frac{1}{10}+c=\frac{1}{5}\Rightarrow c=\frac{1}{10}$

Now, we can write the probabilities for each $x$ using $P(X=x)=\frac{x}{20}+\frac{2}{20}=\frac{x+2}{20}$:
- For $x=0$: $P(X=0)=\frac{2}{20}$
- For $x=1$: $P(X=1)=\frac{3}{20}$
- For $x=2$: $P(X=2)=\frac{4}{20}$
- For $x=3$: $P(X=3)=\frac{5}{20}$
- For $x=4$: $P(X=4)=\frac{6}{20}$

Next, we calculate the expected value of $X^2$:
$E\left[X^2\right]=\sum _{x=0}^4{x}^2·P(X=x)$
$E\left[X^2\right]=0^2\left(\frac{2}{20}\right)+1^2\left(\frac{3}{20}\right)+2^2\left(\frac{4}{20}\right)+3^2\left(\frac{5}{20}\right)+4^2\left(\frac{6}{20}\right)$
$E\left[X^2\right]=\frac{0+3+16+45+96}{20}=\frac{160}{20}=8$

The variance of $X$, denoted by $\alpha$, is:
$\alpha =E\left[X^2\right]-{\left(E\right[X\left]\right)}^2$
$\alpha =8-{\left(\frac{5}{2}\right)}^2=8-\frac{25}{4}=\frac{32-25}{4}=\frac{7}{4}$

Finally, we calculate $24\alpha$:
$24\alpha =24·\frac{7}{4}=6·7=42$