We are given a discrete random variable $X$ that takes the values $-4, -2, -1, 2, 5, 6$ with equal probabilities. We need to compute the variance of $X$, denoted as $Var(X)$, and express the result as an irreducible fraction $p/7$, where $p$ is an integer.

1. Problem Description

We are given a discrete random variable $X$ that takes the values $-4, -2, -1, 2, 5, 6$ with equal probabilities. We need to compute the variance of $X$, denoted as $Var(X)$, and express the result as an irreducible fraction $p/7$, where $p$ is an integer.

2. Solution Steps

Let $X$ be a discrete random variable with possible values $x_i$ and corresponding probabilities $P(X=x_i)$. The variance of $X$ is given by:

$Var(X) = E[X^2] - (E[X])^2$

where $E[X]$ is the expected value of $X$ and $E[X^2]$ is the expected value of $X^2$.

Since the probabilities are equal, $P(X=x_i) = \frac{1}{6}$ for each $x_i$.

First, calculate the expected value $E[X]$:

$E[X] = \sum_{i=1}^{6} x_i P(X=x_i) = \frac{1}{6} (-4 -2 -1 +2 +5 +6) = \frac{1}{6} (6) = 1$

Next, calculate the expected value of $X^2$, $E[X^2]$:

$E[X^2] = \sum_{i=1}^{6} x_i^2 P(X=x_i) = \frac{1}{6} ( (-4)^2 + (-2)^2 + (-1)^2 + 2^2 + 5^2 + 6^2 )$

$E[X^2] = \frac{1}{6} ( 16 + 4 + 1 + 4 + 25 + 36 ) = \frac{1}{6} (86)$

Now, calculate the variance $Var(X)$:

$Var(X) = E[X^2] - (E[X])^2 = \frac{86}{6} - (1)^2 = \frac{43}{3} - 1 = \frac{43}{3} - \frac{3}{3} = \frac{40}{3}$

The problem requires that the answer be in the form $\frac{p}{7}$. However, $\frac{40}{3}$ cannot be expressed in the form $\frac{p}{7}$, since the denominator must be $7$. Perhaps there was an error in transcribing the image.

The values given are $-4, -2, -1, 2, 5, 6$. Since we are asked to represent the variance as $p/7$, we need to double-check if the problem contains other possible numbers.

After close inspection of the handwriting, the problem statement requires us to find integers $p$ and $7$ (note the 7 is explicitly written).

Since the variance is $\frac{40}{3}$, we look for an integer $p$ close to $\frac{40}{3} \times 7 = \frac{280}{3} = 93.333$.

The problem asks us to give the result as an irreducible fraction $p/7$ with integers $p$ and $7$. Since the given result is $\frac{40}{3}$, and we need to transform this into $\frac{p}{7}$, it seems there is a misunderstanding of the requested format.

Going back to the original question: The discrete variable $X$ takes values $-4,-2,-1,2,5,6$. Then $E[X]=1$ and $E[X^2] = \frac{86}{6} = \frac{43}{3}$. $Var(X)=E[X^2]-(E[X])^2=\frac{43}{3}-1 = \frac{40}{3}$.

The question asked to represent the result as $p/7$ with $p$ as an integer. This is an incorrect format. Given the values of X, its impossible to get the form of variance as $p/7$.

3. Final Answer

The variance is $\frac{40}{3}$. It cannot be expressed in the form $p/7$ where $p$ is an integer.

The problem asks us to write the answer as an irreducible fraction $p/7$. Thus, we are seeking an integer $p$. There must be an error. The variance is $\frac{40}{3}$. It cannot be expressed as $p/7$.

There seems to be an error in the instructions.

Final Answer: 40/3