Understanding the Probability Mass Function (PMF) Problem

This question asks us to find the values of constants 'a' and 'b' for a given probability mass function (PMF) of a discrete random variable X. We are provided with the possible values of X (0, 1, 2) and their corresponding probabilities in terms of 'a' and 'b'. Crucially, we are also given that the mean (expected value) of this random variable X is 1.1.

To solve this, we need to use two fundamental properties of a probability mass function:

1. The sum of probabilities for all possible values of the random variable must equal 1.
2. The mean (expected value) of a discrete random variable is the sum of each possible value multiplied by its probability.

Let's look at the provided probability distribution:

Setting Up Equations from PMF Properties

Using the properties mentioned above, we can set up a system of two linear equations with two variables, 'a' and 'b'.

Equation 1: Sum of Probabilities

The sum of all probabilities must be 1:

$\qquad P(X=0) + P(X=1) + P(X=2) = 1$

Substituting the given probabilities in terms of 'a' and 'b':

$\qquad 3a + 3b + 4b = 1$

Combining the 'b' terms, we get our first equation:

$\qquad \text{Equation 1: } 3a + 7b = 1$

Equation 2: Mean of the PMF

The mean ($E(X)$) of a discrete random variable is calculated as $\sum [x \cdot P(X=x)]$. We are given that the mean is 1.1.

$\qquad E(X) = (0 \cdot P(X=0)) + (1 \cdot P(X=1)) + (2 \cdot P(X=2)) = 1.1$

Substituting the given probabilities:

$\qquad (0 \cdot 3a) + (1 \cdot 3b) + (2 \cdot 4b) = 1.1$

Performing the multiplications:

$\qquad 0 + 3b + 8b = 1.1$

Combining the 'b' terms, we get our second equation:

$\qquad \text{Equation 2: } 11b = 1.1$

Solving for 'a' and 'b'

Now we have a system of two linear equations:

1. $3a + 7b = 1$
2. $11b = 1.1$

We can solve Equation 2 for 'b' directly:

$\qquad 11b = 1.1$

Divide both sides by 11:

$\qquad b = \frac{1.1}{11}$

$\qquad b = 0.1$

Now substitute the value of 'b' (0.1) into Equation 1:

$\qquad 3a + 7(0.1) = 1$

$\qquad 3a + 0.7 = 1$

Subtract 0.7 from both sides:

$\qquad 3a = 1 - 0.7$

$\qquad 3a = 0.3$

Divide both sides by 3:

$\qquad a = \frac{0.3}{3}$

$\qquad a = 0.1$

So, the values of a and b are 0.1 and 0.1 respectively.

Verifying the PMF and Mean

Let's check if these values create a valid PMF with the correct mean.

If $a = 0.1$ and $b = 0.1$, the probabilities are:

• $P(X=0) = 3a = 3(0.1) = 0.3$
• $P(X=1) = 3b = 3(0.1) = 0.3$
• $P(X=2) = 4b = 4(0.1) = 0.4$

Check sum of probabilities:

$\qquad 0.3 + 0.3 + 0.4 = 1.0$

The sum is 1, which is correct for a PMF. Also, all probabilities (0.3, 0.3, 0.4) are between 0 and 1, inclusive.

Check the mean:

$\qquad E(X) = (0 \cdot 0.3) + (1 \cdot 0.3) + (2 \cdot 0.4)$

$\qquad E(X) = 0 + 0.3 + 0.8$

$\qquad E(X) = 1.1$

The calculated mean matches the given mean of 1.1.

Thus, the values $a=0.1$ and $b=0.1$ satisfy both conditions.

Conclusion on Finding 'a' and 'b'

By using the fundamental properties of a probability mass function – the sum of probabilities equaling one and the formula for the mean – we were able to set up and solve a system of equations to find the unknown constants 'a' and 'b' in the given probability distribution. The values found are a = 0.1 and b = 0.1.

Revision Table: Probability Mass Function Key Concepts

Additional Information: Properties of Probability Distributions

Probability distributions, like the probability mass function for discrete variables or the probability density function for continuous variables, are essential tools in statistics and probability theory. They describe the likelihood of different outcomes occurring for a random variable.

• Normalization: A key property for any probability distribution is that the total probability across all possible outcomes must sum up to 1 (for discrete) or integrate to 1 (for continuous). This ensures that the variable must take on *some* value within its range.
• Non-negativity: The probability assigned to any single outcome or range of outcomes must be non-negative. Probabilities cannot be negative.
• Expected Value: The mean or expected value represents the center of the distribution. It's a weighted average of the possible outcomes, with the weights being their respective probabilities.
• Variance and Standard Deviation: Beyond the mean, variance and standard deviation measure the spread or dispersion of the distribution. Variance ($Var(X)$) is the expected value of the squared deviation from the mean, and standard deviation ($\sigma$) is the square root of the variance. $Var(X) = E[(X - E(X))^2] = E(X^2) - (E(X))^2$.

Understanding these properties is crucial for working with random variables and their distributions in various statistical applications.