How do I find $\bar{y}$ in this question?

Question

A summary of $20$ observations of $y$ gave the following information

$\sum(y-a)=-37$

$\sum(y-a)^2=1529$

Find the mean and standard deviation of $y$.

In this question, I was able to find the standard deviation like such:

Let $\sum x=\sum(y-a)=-37$

Let $\sum x^2=\sum (y - a)^2=1529$

$\bar{x}=\frac{\sum x}{n}$

$\bar{x}=\frac{-37}{20}$

$\bar{x}=-1.85$

$\sigma_x=\sqrt{\frac{\sum x^2}{n}-(\bar{x})^2}$

$\sigma_x=\sqrt{\frac{1529}{20}-(-1.85)^2}$

$\sigma_x=8.55$

Since $\sigma_y=\sigma_x$

$\sigma_y=8.55$

However, I couldn't progress to solving for the mean.

Also, when I checked the answer, it said that $\bar{y}$ was $-1.85$ which would mean that $a = 0$.

Can someone explain to me how to find $\bar{y}$?

Thanks

Answer 1

A condition is missing for finding the mean $\bar{y}$!

You found $\sigma_y$ correctly by introducing the new variable $x=y-a$: $$\mathbb E(x)=\mathbb E(y-a)=\mathbb E(y)-a \Rightarrow \bar{y}=a+\bar{x}\\ Var(x)=Var(y-a)=Var(y)$$ Alternative direct calculation: $$\sum(y-a)=-37\Rightarrow \sum y=20a-37 \Rightarrow \bar{y}=a-1.85;\\ \sigma_y=\sqrt{\frac{\sum(y-\bar{y})^2}{20}}=\sqrt{\frac{\sum(y-a+1.85)^2}{20}}=\sqrt{\frac{\sum(y-a)^2+2\cdot 1.85\sum(y-a)+1.85^2}{20}}=\\ \sqrt{\frac{1529+3.7\cdot (-37)+20\cdot 1.85^2}{20}}=8.5456....$$

Answer 2

The question depending on $a$. You can see this if you introduce $y = z - \Delta a$ and change $a$ to $\tilde{a}-\Delta a$ you will see that the equation will not change. Hence, you need to express the mean by rewriting the first equation. Divide the equation by the number of observations $20$ to obtain

$$\bar{y}-a=-37/20 \implies \bar{y}=a-37/20$$

Now, rewrite the second equation as

$$\sum_{n=1}^{20}\left[y^2-2ay+a^2\right] = 1529$$ divide by $20$ $$\implies \bar{y^2}-2a\bar{y}+a^2 = 1529/20$$ solve for $\bar{y^2}$ and use $\bar{y}=a-37/20$ $$\bar{y^2}=2a(a-37/20)-a^2+1529/20$$

The standard deviation is given by

$$\sigma_Y=\sqrt{\bar{y^2}-\bar{y}^2}=\sqrt{2a(a-37/20)-a^2+1529/20-(a-37/20)^2}$$