I am stumped on a question. It gives me the mean value as 24, standard deviation is 4 and N = 10. What is the sum of (x-mean)^2
What I've tried is as follow
$$ \text{Standard Deviation} = \sqrt{\dfrac{\sum (x - x')^{2}}{N}} $$
$$ 4 = \sqrt{\dfrac{\sum_{}^{}(x - x')^{2}}{10}} $$
$$ \sum_{}^{}(x - x')^{2} = 160 $$
Is that an acceptable answer?
It can be shown that $\sigma^2_X = \mu_{X^2} - \mu^2_{X}$. Note that the definition of variance is the sum given. You're summing the squared deviations from the mean, which is part of computing variance. Once you know the standard deviation (which is given to you), you have the variance for free.