If dataset $A=\{1,2,3,\ldots,19\}$ and dataset $B=\{ax+b;\,x\in A\}$. If mean of $B$ is $30$ and variance of $B$ is $750$, then sum of possible values of $b$ is

Question

If the mean deviation about the median of the numbers \[ k,\,2k,\,3k,\,\ldots,\,1000k \] is $500$, then $k^2$ is equal to

Answer 1

To solve this problem, we need to analyze the datasets $ A $ and $ B $ and use the given statistical properties.

First, understand the datasets:
- Dataset $ A = \{1, 2, 3, \ldots, 19\} $.
- Dataset $ B = \{ax+b; \, x\in A\} $ means $ B $ is a linear transformation of $ A $ with slope $ a $ and intercept $ b $.
Calculate the mean of dataset $ A $:
- The mean of an arithmetic sequence is the average of the first and last term. So,
- $\text{Mean of } A = \frac{1 + 19}{2} = 10$.
Given the mean of $ B $ is 30, set up the equation:
- The mean of $ B $ is given by substituting the mean of $ A $ into the transformation:
- $a \times 10 + b = 30$, simplifying to $10a + b = 30$. (Equation 1)
Calculate the variance of dataset $ A $:
- The variance of an arithmetic sequence from 1 to $ n $ is $\frac{(n^2-1)}{12}$.
- For $ A $, where $ n = 19 $, $\text{Variance of } A = \frac{(19^2 - 1)}{12} = \frac{360}{12} = 30$.
Given the variance of $ B $ is 750, relate it with $ a^2 \times \text{Variance of } A $:
- The variance of $ B $ is transformed as $ a^2 \times \text{Variance of } A = 750 $.
- $a^2 \times 30 = 750$, simplifying to $a^2 = 25$.
- Solve for $ a $: $a = 5$ or $a = -5$.
Find possible values of $ b $ using $ a = 5 $ and $ a = -5 $:
- For $ a = 5 $, substitute into Equation 1: $10 \times 5 + b = 30$ giving $b = -20$.
- For $ a = -5 $, substitute into Equation 1: $10 \times (-5) + b = 30$ giving $b = 80$.
Sum the possible values of $ b $:
- Possible values of $ b $ are $-20$ and $80$.
- The sum of these values is $-20 + 80 = 60$.

Conclusion: The sum of possible values of $ b $ is 60.

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