Descriptive Statistics

Mean: $(2+4+4+6+9)/5 = 25/5 = 5$. Median (middle of the sorted set): $4$. So $m + M = 5 + 4 = 9$.

Written in increasing order, the six daily totals are:

$$6,\ 8,\ 10,\ 11,\ 15,\ 19$$

With $n = 6$ there is no single central day: the two central positions are the $3$rd and the $4$th, holding $10$ and $11$.

The median is the mean of that pair:

$$\dfrac{10 + 11}{2} = 10.5$$

The frequencies are $3$, $7$, $5$, $4$ and $1$, and they total $3 + 7 + 5 + 4 + 1 = 20$, matching the $20$ matches played.

The largest frequency is $7$, and it sits in the column headed $1$: the team scored exactly $1$ goal in $7$ matches, more often than any other score. No other score is reached that many times, so the modal value is unique.

$$\text{mode} = 1$$

Mean: $\bar{x} = \dfrac{1+2+3}{3} = 2$.

Squared deviations:

$$(1-2)^2 + (2-2)^2 + (3-2)^2 = 1 + 0 + 1 = 2$$

So $\sigma^2 = \dfrac{2}{3}$.

By definition, the standard deviation $\sigma$ is the square root of the variance $\sigma^2$:

$$\sigma = \sqrt{\sigma^2}$$

The mean and the sum divided by $n$ denote the arithmetic mean, not the standard deviation.

With $n=3$ values:

$$\bar{x} = \dfrac{2 + 4 + 6}{3} = \dfrac{12}{3} = 4$$

Mean: $\bar{x} = \dfrac{4+4+4+4}{4} = 4$.

All deviations are zero: $x_i - \bar{x} = 0$ for every $i$, so

$$\sigma^2 = \dfrac{1}{4}\bigl(0^2+0^2+0^2+0^2\bigr) = 0$$

Sum $= 2 + 4 + 4 + 6 + 9 = 25$.

Mean $= \dfrac{25}{5} = 5$.

The eight original numbers add up to:

$$8 \times 12 = 96$$

The seven numbers that remain add up to:

$$7 \times 13 = 91$$

The two totals contain exactly the same numbers except for the one that was deleted, so that number is the gap between them:

$$96 - 91 = 5$$

As a check, $5$ lies below the original mean of $12$, so deleting it pulls the mean of what is left upwards — consistent with the new mean of $13$.

Total frequency: $\sum f = 1+2+4+2+1 = 10$.

Weighted total: $\sum f x = 0\cdot 1 + 1\cdot 2 + 2\cdot 4 + 3\cdot 2 + 4\cdot 1 = 0+2+8+6+4 = 20$, so

$$\bar{x} = \dfrac{20}{10} = 2$$

Weighted squared deviations, using $d = x - 2$:

$$\sum f\,(x-\bar{x})^{2} = 1\cdot 4 + 2\cdot 1 + 4\cdot 0 + 2\cdot 1 + 1\cdot 4 = 12$$

Dividing by the total frequency $10$:

$$\sigma^{2} = \dfrac{12}{10} = 1.2$$