1. Set up the probability mass function

Since the points $\bigl(x, P(X=x)\bigr)$ for $x=0,1,2,3,4$ lie on a straight line, let

$P(X=x)=ax+b, \qquad x=0,1,2,3,4.$

Also, $P(X=x)=0$ for all other real $x$.

So the probabilities are:

$P(0)=b,\quad P(1)=a+b,\quad P(2)=2a+b,\quad P(3)=3a+b,\quad P(4)=4a+b.$

2. Use the total probability condition

Since probabilities sum to $1$,

$\sum_{x=0}^4 P(X=x)=1.$

Thus,

$b+(a+b)+(2a+b)+(3a+b)+(4a+b)=1.$

10a+5b=1. \tag{1}

3. Use the mean condition

Given $E(X)=\dfrac{5}{2}$,

$\sum_{x=0}^4 xP(X=x)=\frac{5}{2}.$

So,

$0\cdot b+1(a+b)+2(2a+b)+3(3a+b)+4(4a+b)=\frac{5}{2}.$

Simplifying,

$(1+4+9+16)a + (1+2+3+4)b = \frac{5}{2}$

30a+10b=\frac{5}{2}. \tag{2}

Multiply (2) by $2$:

60a+20b=5 \implies 6a+2b=\frac12. \tag{3}

From (1):

10a+5b=1 \implies 2a+b=\frac15. \tag{4}

Now subtract twice (4) from (3):

$6a+2b-(4a+2b)=\frac12-\frac25$

$2a=\frac{1}{10} \implies a=\frac{1}{20}.$

Then from (4):

$2\left(\frac{1}{20}\right)+b=\frac15$

$\frac{1}{10}+b=\frac15 \implies b=\frac{1}{10}.$

Hence,

$P(X=x)=\frac{x}{20}+\frac{1}{10} = \frac{x+2}{20}, \qquad x=0,1,2,3,4.$

So,

$P(0)=\frac{2}{20},\; P(1)=\frac{3}{20},\; P(2)=\frac{4}{20},\; P(3)=\frac{5}{20},\; P(4)=\frac{6}{20}.$

4. Compute $E(X^2)$

$E(X^2)=\sum_{x=0}^4 x^2 P(X=x).$

Thus,

$E(X^2)=0^2\cdot\frac{2}{20}+1^2\cdot\frac{3}{20}+2^2\cdot\frac{4}{20}+3^2\cdot\frac{5}{20}+4^2\cdot\frac{6}{20}.$

$E(X^2)=0+\frac{3}{20}+\frac{16}{20}+\frac{45}{20}+\frac{96}{20}.$

$E(X^2)=\frac{160}{20}=8.$

5. Compute the variance

$\operatorname{Var}(X)=E(X^2)-[E(X)]^2.$

So,

$\alpha=8-\left(\frac52\right)^2=8-\frac{25}{4}=\frac{32-25}{4}=\frac{7}{4}.$

6. Find $24\alpha$

$24\alpha=24\cdot\frac74=6\cdot 7=42.$

Final Answer

$24\alpha=42$