SAT Question f8a322d9: Distributions, Center & Spread answer and explanation with Desmos solution

Choice B is correct. The histograms shown have the same shape, but data set A contains values between $20$ and $60$ and data set B contains values between $10$ and $50$. Thus, the mean of data set A is greater than the mean of data set B. Therefore, the smallest possible difference between the mean of data set A and the mean of data set B is the difference between the smallest possible mean of data set A and the greatest possible mean of data set B. In data set A, since there are $3$ integers in the interval greater than or equal to $20$ but less than $30$, $4$ integers greater than or equal to $30$ but less than $40$, $7$ integers greater than or equal to $40$ but less than $50$, and $9$ integers greater than or equal to $50$ but less than $60$, the smallest possible mean for data set A is $\frac{\left(3\cdot 20\right)+\left(4\cdot 30\right)+\left(7\cdot 40\right)+\left(9\cdot 50\right)}{23}$. In data set B, since there are $3$ integers greater than or equal to $10$ but less than $20$, $4$ integers greater than or equal to $20$ but less than $30$, $7$ integers greater than or equal to $30$ but less than $40$, and $9$ integers greater than or equal to $40$ but less than $50$, the largest possible mean for data set B is $\frac{\left(3\cdot 19\right)+\left(4\cdot 29\right)+\left(7\cdot 39\right)+\left(9\cdot 49\right)}{23}$. Therefore, the smallest possible difference between the mean of data set A and the mean of data set B is $\frac{\left(3\cdot 20\right)+\left(4\cdot 30\right)+\left(7\cdot 40\right)+\left(9\cdot 50\right)}{23}-\frac{\left(3\cdot 19\right)+\left(4\cdot 29\right)+\left(7\cdot 39\right)+\left(9\cdot 49\right)}{23}$, which is equivalent to $\frac{\left(3\cdot 20\right)-\left(3\cdot 19\right)+\left(4\cdot 30\right)-\left(4\cdot 29\right)+\left(7\cdot 40\right)-\left(7\cdot 39\right)+\left(9\cdot 50\right)-\left(9\cdot 49\right)}{23}$. This expression can be rewritten as $\frac{3\left(20-19\right)+4\left(30-29\right)+7\left(40-39\right)+9\left(50-49\right)}{23}$, or $\frac{23}{23}$, which is equal to $1$. Therefore, the smallest possible difference between the mean of data set A and the mean of data set B is $1$.