IF the mean of a set of scores is 80 and standard deviation is 10, explain the effect on mean and standard deviation in case following changes are made to each of the scores
1) adding 5
2) multiplying by 5
closed as off-topic by Sally, Davide Giraudo, Ken, Carl Mummert, avid19 Sep 6 '15 at 1:45
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There are two types of explanation, formal and informal. We will try to do both.
Let $X$ be a random variable, and let $a$ be a constant. Let $Y=X+a$. then $E(Y)=E(X+a)=E(X)+E(a)=E(X)+a$. So adding the constant $a$ adds $a$ to the mean.
The standard deviation of a random variable $W$ is defined to be the square root of the variance. If $\mu_W$ is the mean of $W$, then the variance of $W$ is defined to be $E(W-\mu_W)^2$.
So the variance of $X$ is $E(X-\mu_X)^2$.
The variance of $Y$ is $E(Y-\mu_Y)^2$.
But $Y=X+a$ and $\mu_Y=\mu_X+a$. So $Y-\mu_Y=(X+a)-(\mu_X+a)=X-\mu_X$, and therefore the variance of $Y$ is equal to the variance of $X$.
Informally, if the mean score in a class is $60$, and we add $5$ to everybody's score, then the mean score (class average) should increase by $5$. However, the standard deviation measures the "wiggliness" of the scores, their variability, and that has not changed. The standard deviation is a kind of measure of the average distance from the mean. since all scores and the mean have changed by the same amount, the average distance from the mean has not changed.
For multiplying by $5$, there is a formal argument similar in structure to the one for sum. Let $Y=kX$. We get $E(Y)=E(kX)=kE(X)$. So the mean gets multiplied by $k$.
There is a similar argument for variance. If $Y=kx$, then $\mu_Y=k\mu_X$.
So $E(Y-\mu_Y)^2=E(kX-k\mu_X)^2=k^2E(X-\mu_X)^2$. So the variance gets multiplied by $k^2$, and therefore the standard deviation gets multiplied by $|k|$.
For an informal explanation, suppose that instead of giving the score out of $100$, we decide to do it out of $500$, by multiplying each person's score by $5$. Then all marks are out of $500$. The class average will also get multiplied by $5$.
Standard deviation is less clear. But it seems plausible that the wiggliness of the scores increases by a factor of $5$, since for example a difference between Alicia and Bob which used to be $8$ has become $60$. Remember that standard deviation is a kind of measure of average departure from the mean. Scores and mean have been multiplied by $5$, so average distance from the mean has also been multiplied by $5$.