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

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: $$\sum(y-a)=-37\Rightarrow \sum y=20a-37 \Rightarrow \bar{y}=a-1.85;\\ \sigma_y=\sqrt{\frac{\sum(y-\bar{y})^2}{20}}=\sqrt{\frac{\sum(y-a+1.85)^2}{20}}=\sqrt{\frac{\sum(y-a)^2+2\cdot 1.85\sum(y-a)+1.85^2}{20}}=\\ \sqrt{\frac{1529+3.7\cdot (-37)+20\cdot 1.85^2}{20}}=8.5456....$$

• Thank you! I thought I was forgetting something in my working. – ianc1339 Mar 30 at 21:14
• You are welcome. Good luck. – farruhota Mar 31 at 0:39

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

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

• Thank you for your answer. However, I am still not sure how to find $\bar{y}$ as mentioned in my question. Is it just that there isn't enough information given in the question to find $\bar{y}$? – ianc1339 Mar 30 at 9:28
• I mean with the information given so far, I guess it just means that $a$ can be any real numbers. Thus, there isn't enough information to find $\bar{y}$ perhaps? – ianc1339 Mar 30 at 9:30
• Thus, the standard deviation does not depend on $a$ – Claude Leibovici Mar 30 at 9:37
• Since the mean depends on $a$, how would I find $\bar{y}$? – ianc1339 Mar 30 at 9:40
• @ianc1339 We only can say that $\bar{y}=a-37/20$ for $a\in \mathbb{R}$. As no further information is provided we cannot say more about $\bar{y}$. – MachineLearner Mar 30 at 14:17