The standard deviation of $\mathrm{n}$ numbers $x_1, x_2, x_3, \ldots \ldots ., x_n$, with mean $\mathrm{x}$ is equal to $\sqrt{\frac{s}{n}}$, where $\mathrm{S}$ is the sum of the squared differences, $\left(x_i-x\right)^2$ for $1 \leq \mathrm{i} \leq \mathrm{n}$. If the standard deviation of the 4 numbers 140 -a, 140,160 , and $160+\mathrm{a}$ is 50 , where $\mathrm{a}>0$, what is the value of a?
The mean is 150.
$s=(140-a-150)^2+(140-150)^2+(160-150)^2+(160+a-150)^2=(-10-a)^2+(-10)^2+(10)^2+(10+a)^2=2(100+20a+a^2)+2(100).$
$\sqrt{\dfrac{s}{n}}=\dfrac{1}{2}\sqrt{2(100+20a+a^2+100)}=50$
$2\sqrt{\dfrac{s}{n}}=100=\sqrt{(2)(200+20a+a^2)}$
There's no real solution for a.
If you know the standard deviation is $50$ then you know that the variance is $50^2 = 2500$, which is $s/4$:
$$s/4 = 2500 = (10+a)^2/2 + 50.$$
Then $(10+a)^2 = 4900$ and you quickly arrive at $a = 60$.