Descriptive Statistics practice questions — A-Level Maths

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.

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:

$${\displaystyle \frac{10+11}{2}}=10.5$$

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

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

$${\sigma }^2={\displaystyle \frac{1}{4}}(0^2+0^2+0^2+0^2)=0$$

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

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

Mean $={\displaystyle \frac{25}{5}}=5$.

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.

$$\text{mode}=1$$

With $n=3$ values:

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

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

Squared deviations:

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

So ${\sigma }^2={\displaystyle \frac{2}{3}}$.

Original mean: $\bar{x}={\displaystyle \frac{4+8+10+12+16}{5}}={\displaystyle \frac{50}{5}}=10$.

Deviations $-6,-2,0,2,6$ give

$${\sigma }^2={\displaystyle \frac{36+4+0+4+36}{5}}={\displaystyle \frac{80}{5}}=16,\sigma =4$$

The new list is $0.5,2.5,3.5,4.5,6.5$, with mean ${\displaystyle \frac{17.5}{5}}=3.5$. Its deviations are $-3,-1,0,1,3$, so

$${\sigma }_{\text{new}}^2={\displaystyle \frac{9+1+0+1+9}{5}}=4$$

Subtracting $3$ moved the mean by $3$ as well, so every deviation was unchanged; halving each value halved every deviation, so the standard deviation is halved:

$${\sigma }_{\text{new}}={\displaystyle \frac{4}{2}}=2$$

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 $13$.