Suppose $a ≤ x_i≤ b $ for $i = 1, . . . , n. $ Show that $a ≤ \bar x ≤ b.$

Question

Question is essentially, suppose a data set $x_i$ has values between (or equal to) $a$ and $b$, show that the mean lies in the same interval.

Obviously the mean lies inside the range however I'm struggling to 'show' this. Tried a few different routes with considering the summation of $x_i$ for $\bar x$ but not actually got anything valid.

Please help!

Answer 1

If $x_i\le b$ for $i=1,\ldots,n$, then $x_1+\ldots+x_n\le b+\ldots+b = nb$, so $$ \bar x = \frac{x_1+\ldots+x_n}{n}\,\le\,\frac{nb}{n} = b. $$ Can you do the same for the lower bound?

Answer 2

Hint :

Each element $x_{i}$ can be written as $$x_i = a + c_i$$

For some constant $c_i$

Similarly Each element $x_{i}$ can also be written as $$x_i = b - d_i$$

For some constant $d_i$

Can you now find the upper and lower limit of these summation ??

Answer 3

Let's see. I'll do the lower bound, since the upper bound can be done in a similar fashion. Since $a\leq x_i$ for all $1\leq i\leq n$, we have $$a+a+\cdots+a\leq x_1+x_2+\cdots+x_n,$$

where I have added up each of the inequalities $a\leq x_i$.

The left-hand-side of this inequality is $na$, so the inequality is the same as

$$na \leq x_1+x_2+\cdots +x_n.$$

If I divide both sides of this inequality by $n$, I get $$a \leq \frac{x_1+x_2+\cdots +x_n}{n},$$

but the right-hand-side is exactly the mean $\bar{x}$ of the data points, so in total

$$a\leq\bar{x}.$$