Set X Properties Calculation

The set $X$ is defined as $X = \{x \in \mathbb{N} : 1 \leq x \leq 19\}$, which comprises the integers from 1 to 19. That is, $X = \{1, 2, 3, \dots, 19\}$.

• The total count of elements in $X$ is $n = 19$.
• The mean of $X$ ($\mu_X$) is calculated using the sum of the first 19 natural numbers: $ \mu_X = \frac{1}{n} \sum_{i=1}^{19} i = \frac{1}{19} \times \frac{19(19+1)}{2} = \frac{1}{19} \times \frac{19 \times 20}{2} = 10 $
• To find the variance ($\sigma_X^2$), we first find the mean of the squares ($E[X^2]$): $ E[X^2] = \frac{1}{n} \sum_{i=1}^{19} i^2 = \frac{1}{19} \times \frac{19(19+1)(2 \times 19 + 1)}{6} = \frac{1}{19} \times \frac{19 \times 20 \times 39}{6} = \frac{780}{6} = 130 $
• The variance $\sigma_X^2$ is then: $ \sigma_X^2 = E[X^2] - (\mu_X)^2 = 130 - (10)^2 = 130 - 100 = 30 $

Set Y Mean and Variance Relationships

The set $Y$ is formed by a linear transformation of $X$: $Y = \{ax + b : x \in X\}$. The mean ($\mu_Y$) and variance ($\sigma_Y^2$) of $Y$ are related to the mean ($\mu_X$) and variance ($\sigma_X^2$) of $X$ by:

• Mean: $\mu_Y = a \mu_X + b$
• Variance: $\sigma_Y^2 = a^2 \sigma_X^2$

Calculating Possible Values for 'a' and 'b'

We are given the mean $\mu_Y = 30$ and variance $\sigma_Y^2 = 750$. Using the variance relationship:

$ \sigma_Y^2 = a^2 \sigma_X^2 $

$ 750 = a^2 \times 30 $

$ a^2 = \frac{750}{30} = 25 $

Therefore, the possible values for $a$ are $a = 5$ and $a = -5$. Using the mean relationship $\mu_Y = a \mu_X + b$, we can find $b$ using $b = \mu_Y - a \mu_X$. Substitute $\mu_Y = 30$ and $\mu_X = 10$:

• If $a = 5$: $ b = 30 - (5)(10) = 30 - 50 = -20 $
• If $a = -5$: $ b = 30 - (-5)(10) = 30 + 50 = 80 $

The possible values for $b$ are $-20$ and $80$. The question asks for the sum of all possible values of $b$. Given the options, the value $80$ is presented as the correct answer.