a)

The mean is the sum of all the responses ($\sum x$) divided by the number of responses ($n$).

Here, $n = 20$.

\begin{align}
\sum x &= \var{a[0]} + \var{a[1]} +\var{a[2]} +\var{a[3]} +\var{a[4]} +\var{a[5]} +\var{a[6]} +\var{a[7]} +\var{a[8]} +\var{a[9]} + \var{a[10]} + \var{a[11]} +\var{a[12]} +\var{a[13]} +\var{a[14]} +\var{a[15]} +\var{a[16]} +\var{a[17]} +\var{a[18]} +\var{a[19]} \\
&= \var{sum(a)} \text{.}
\end{align}

Therefore we calculate the mean

\begin{align}
\overline{x} &= \frac{\sum x}{n} \\[0.5em]
&= \frac{\var{sum(a)}}{20} \\[0.5em]
&= \var{mean} \text{.}
\end{align}

b)

The median is the middle value. We need to sort the list in order:

\[ \var{a_s[0]}, \quad \var{a_s[1]}, \quad \var{a_s[2]}, \quad \var{a_s[3]}, \quad \var{a_s[4]}, \quad \var{a_s[5]}, \quad \var{a_s[6]}, \quad \var{a_s[7]}, \quad \var{a_s[8]}, \quad \var{a_s[9]}, \quad \var{a_s[10]}, \quad \var{a_s[11]}, \quad \var{a_s[12]}, \quad \var{a_s[13]}, \quad \var{a_s[14]}, \quad \var{a_s[15]}, \quad \var{a_s[16]}, \quad \var{a_s[17]}, \quad \var{a_s[18]}, \quad \var{a_s[19]} \]

There is an even number of responses, so there are two numbers in the middle (10^th and 11^th place). To find the median, we need to find the mean of these two numbers $\var{a_s[9]}$ and $\var{a_s[10]}$:

\begin{align}
\frac{\var{a_s[9]} + \var{a_s[10]}}{2} &=  \frac{\var{a_s[9] + a_s[10]}}{2} \\
&= \var{median} \text{.} 
\end{align}

c)

The mode is the value that occurs the most often in the data.

To find a mode, we can look at our sorted list:

$\var{a_s[0]}, \var{a_s[1]}, \var{a_s[2]}, \var{a_s[3]}, \var{a_s[4]}, \var{a_s[5]}, \var{a_s[6]}, \var{a_s[7]}, \var{a_s[8]}, \var{a_s[9]}, \var{a_s[10]}, \var{a_s[11]}, \var{a_s[12]}, \var{a_s[13]}, \var{a_s[14]}, \var{a_s[15]}, \var{a_s[16]}, \var{a_s[17]}, \var{a_s[18]}, \var{a_s[19]}$.

We notice that $\var{mode1}$ occurs the most ($\var{modetimes[mode1]}$ times) so $\var{mode1}$ is the mode.

d)

Range is the difference between the highest and the lowest value in the data.

To find this, we subtract the lowest value from the highest value:

\[ \var{max(a)} - \var{min(a)} = \var{range} \text{.}\]

a)

We calculate the midpoint by finding the mean of the lower and upper bounds. For example, the midpoint for $40 \leq a \lt 50$ is

\[\begin{align} \displaystyle \frac{40 + 50}{2} &= \frac{90}{2} \\&= 45 \text{.} \end{align}\]

To calculate the final column, we multiply the second and the third columns:

\[\var{freq[4]} \times 45 = \var{45*freq[4]} \text{.} \]

Finally, when we have completed all the other values, we can calculate the total for the last column.

\[\begin{align} \text{Total} &= 5\times\var{freq[0]} + 15\times\var{freq[1]} + 25\times\var{freq[2]} + 35\times\var{freq[3]} + 45\times\var{freq[4]} + 55\times\var{freq[5]} + 65\times\var{freq[6]} + 75\times\var{freq[7]} + 85\times\var{freq[8]} + 95\times\var{freq[9]} \\&= \var{5*freq[0]} + \var{15*freq[1]} + \var{25*freq[2]} + \var{35*freq[3]} + \var{45*freq[4]} + \var{55*freq[5]} + \var{65*freq[6]} + \var{75*freq[7]} + \var{85*freq[8]} + \var{95*freq[9]}  \\&= \var{freq_midpoint} \text{.} \end{align}\]

The completed frequency table looks like this:

b)

We don't know exactly how much each student got, so we have to assume all students got the midpoint of their group. To estimate the sum of all student grades in a particular interval, we can multiply the interval's midpoint by its frequency.

The sum of all the grades in the class is the sum of the estimates for each interval. We've already calculated this in the table above.

So the estimate for the mean, $\bar{a}$, is as follows:

\begin{align}
\bar{a} &\approx \frac{\var{freq_midpoint}}{\var{sum(freq)}} \\
&= \var{freq_midpoint/sum(freq)} \text{.}
\end{align}

Rounding the mean to 2 decimal places, we get $\var{precround(freq_midpoint/sum(freq),2)}$.

c)

The modal class is the interval with the highest frequency. In this case, the interval $50 \leq a \lt 60$ is the modal class.

a)

Organising the data in a frequency table helps to make mistakes less likely when calculating statistics from our data, summarising the responses all in one place with fewer numbers.

Each row of the frequency column gives the number of students with the corresponding number of siblings.

Always remember to check whether your frequency column adds up to the total (here, it is $30$) to make sure you have not left out any responses.

b)

Mean

The mean number of siblings is the total number of siblings, $\sum x$, divided by the number of students in the sample, $n$.

\begin{align}  
\sum x &= 0 \times \var{freq[0]} + 1 \times \var{freq[1]} + 2 \times \var{freq[2]} + 3 \times \var{freq[3]} + 4 \times \var{freq[4]} + 5 \times \var{freq[5]} + 6 \times \var{freq[6]}
\\
&= 0 + \var{1*freq[1]} + \var{2*freq[2]} + \var{3*freq[3]} + \var{4*freq[4]} + \var{5*freq[5]} + \var{6*freq[6]} \\&= \var{sum(a)} \text{.}
\end{align}

The total number of students $n$ is $30$.

Therefore the mean is

\begin{align}
\bar{x} &= \frac{\sum x}{n} \\
&= \frac{\var{sum(a)}}{30} \\
&= \var{mean} \text{.}
\end{align}

Mode

The mode is the value with the highest frequency. Here, the mode is $\var{mode}$ siblings, with frequency $\var{freq[mode]}$.

Median

The median is the "middle" value in the sample, when arranged in numerical order.

Since $n = 30$, we have two middle values in this data (15^th and 16^th place). We can count from the top of the table until we locate rows where these middle values lie, as the numbers in the table are already sorted by order.

Here, both $15$^th and $16$^th value lie in the row $\var{asa[14]}$.

As $15$^th value $= 16$^th value $= \var{asa[14]}$, the median is $\var{asa[14]}$.

We denote Sweet Heaven as $s$ and Tasty Hell as $t$.

a)

We are going to start with completing the column for Sweet Heaven.

First, we need to find the sum of weights of all the scoops:

\[\begin{align}  \sum s &= \var{a[0]} + \var{a[1]} + \var{a[2]} + \var{a[3]} + \var{a[4]} + \var{a[5]} + \var{a[6]} + \var{a[7]} + \var{a[8]} + \var{a[9]}   \\&= \var{suma} \text{.}
\end{align}\]

The total number of measurements $n$ is $10$.

Therefore the mean is

\[ \begin{align} \overline{s} &= \frac{\sum s}{n} \\[3pt]&= \frac{\var{suma}}{10} \\&= \var{meana} \text{.} \end{align}\]

The median is the middle value. We need to sort the list in order:

There is an even number of responses, so there are two numbers in the middle (5^th and 6^th place). To find the median, we need to find the mean of these two numbers $\var{asort[4]}$ and $\var{asort[5]}$:

\[ \displaystyle \begin{align} \frac{\var{asort[4]} + \var{asort[5]}}{2} &=  \frac{\var{asort[4] + asort[5]}}{2} \\&= \var{mediana} \text{.} \end{align}\]

The mode is the value that occurs the most often in the data.

To find a mode, we can look at our sorted list:

We notice that $\var{modea}$ occurs the most times (3) and so $\var{modea}$ is the mode.

The range is the difference between the highest and the lowest value in the data.

To find this, we subtract the lowest value from the highest value:

\[ \var{max(a)} - \var{min(a)} = \var{rangea} \text{.}\]

So the first column is

Similarly for Tasty Hell,

\[\begin{align}  \sum t &= \var{b[0]} + \var{b[1]} + \var{b[2]} + \var{b[3]} + \var{b[4]} + \var{b[5]} + \var{b[6]} + \var{b[7]} + \var{b[8]} + \var{b[9]}   \\&= \var{sumb} \text{.}
\end{align}\]

The total number of measurements $n$ is $10$ again.

Therefore the mean is

\[ \begin{align} \overline{t} &= \frac{\sum t}{n} \\[3pt]&= \frac{\var{sumb}}{10} \\&= \var{meanb} \text{.} \end{align}\]

For median, we sort the list in order:

There is an even number of responses, so there are two numbers in the middle (5^th and 6^th place). We find the mean of these two numbers $\var{bsort[4]}$ and $\var{bsort[5]}$:

\[ \displaystyle \begin{align} \frac{\var{bsort[4]} + \var{bsort[5]}}{2} &=  \frac{\var{bsort[4] + bsort[5]}}{2} \\&= \var{medianb} \text{.} \end{align}\]

For mode, we look at our sorted list:

We notice that $\var{modeb}$ occurs the most times (2) and so $\var{modeb}$ is the mode.

To find the range, we subtract the lowest value from the highest value:

\[ \var{max(b)} - \var{min(b)} = \var{rangeb} \text{.}\]

So the complete table is​

Let's look at the differences between the two ice cream parlours:

The range of weight of Tasty Hell scoops ($\var{rangeb}$) is far greater than that of Sweet Heaven scoops ($\var{rangea}$).

The mean weight for each shop is $\var{meanab}$. This implies that the scoops are more-or-less the same in both shops. However, looking at the actual values as well as other measures, we can see this is not true, so the mean is not very reliable in this case.

When we compare the medians ($\var{mediana}$ and $\var{medianb}$), we might assume that the scoops are generally lighter in Tasty Hell. This is partly true, but there were some much heavier scoops provided by this shop as well.

Looking at modes ($\var{modea}$ and $\var{modeb}$) can be very misleading, because the modal weight for Tasty Hell is the maximum value at the same time, so it is not a reliable measure of average in this case.

b)

Alice wants her children's ice creams to be very similar.

This is more likely to happen in the shop with a lower range of values.

Comparing the ranges, the range of weight of Sweet Heaven scoops ($\var{rangea}$) is far lower than that of Tasty Hell scoops ($\var{rangeb}$), implying Sweet Heaven is more consistent with their scoops.

The relative frequency of an outcome is the frequency of the outcome divided by the number of trials.

a)

We are told that $\var{no_people}$ people were asked whether they preferred to buy free-range eggs or caged eggs in supermarkets and that $\var{free_range}$ of these people said that they preferred to buy free-range eggs. 

To calculate the relative frequency of people who prefer buying free-range eggs we need the number of trials and the frequency of people who said that they preferred buying free-range eggs.

So, the number of trials in this situation is the number of people who were asked the question, which is $\var{no_people}$.

The frequency of people who said that they preferred to buy free-range eggs is $\var{free_range}$.

Therefore, the relative frequency of people who prefer buying free-range eggs is 

\[
\frac{\var{free_range}}{\var{no_people}} = \var{dpformat({free_range/no_people}, 2)} \; (\text{rounded to $2$ decimal places}).
\] 

b)

We are told that the relative frequency of a student being taller than $150$ cm is $\var{rel_freq}$.

Here, we must use the formula for relative frequency in reverse in order to estimate the number of students in the class who are taller than $150$ cm.

As we are using relative frequency to calculate this number, our answer may not be completely accurate, therefore our answer will be an estimate of the actual number.

If we let $n$ denote the number of students in the class who are taller than $150$ cm and if there are $\var{no_students}$ students in the class then

\[
\begin{align}
\frac{n}{\var{no_students}} &= \var{rel_freq}\\
n &= \var{rel_freq} \times \var{no_students}\\
&= \var{{rel_freq}*{no_students}}.
\end{align}
\]

As $n$ represents a number of people we must round our value of $n$ to the nearest integer.

So, the estimated number of students in the class who are taller than $150$ cm is $\var{dpformat({rel_freq}*{no_students},0)}$.

c)

i) 

Using the frequency table given in the question, we can calculate the sample size of the survey by adding together the frequencies of each of the different types of pets.

So, the sample size of the survey is 

\[
\var{dog}+\var{cat}+\var{hamster}+\var{parrot} = \var{{dog}+{cat}+{hamster}+{parrot}}.
\]

ii) 

To calculate the relative frequency of a person having a dog as a pet, we divide the frequency of people in the survey who had a dog as a pet by the sample size of the survey.

So, the relative frequency is 

\[
\frac{\var{dog}}{\var{n}} = \var{dpformat({dog/n}, 2)} \; (\text{rounded to $2$ decimal places}).
\]

If we are given the probability of an event occurring in a single trial then we can calculate the expected number of times that this event would occur in a larger number of trials.

To do this, we multiply the probability of the event occurring in a single trial by the total number of trials:

\[\text{Expected number of times an event occurs} = \text{Probability of event} \times \text{Number of trials}.\] 

We are given the probabilities that someone buys a ticket to see each film in the table below.

We are also told that $\var{no_people}$ people each buy a ticket at the cinema to see a film of their own choosing during this day.

a)

To calculate the expected number of people who bought tickets to see one of these films we multiply the probability that a person buys a ticket for that film by how many people bought tickets for a film at the cinema. 

So the expected number of people who bought tickets to see Forgotten Game is

\[
\var{Avatar} \times \var{no_people} = \var{{Avatar}*{no_people}}.
\]

b)

We are now asked to calculate the expected number of people who bought tickets to see a Sci-Fi film.

From the table above we can see that there are two films which belong to the Sci-Fi genre: Forgotten Game and The Diamond Valley.

Firstly, we need to calculate the probability that a person buys a ticket to see a Sci-Fi film, which we will denote $P(\text{Sci-Fi})$.

Since the probability that a person buys a ticket to see each film is different, it would be incorrect to say that the probability that a person buys a ticket to see a Sci-Fi film is 

\[\displaystyle\frac{2}{4} = \displaystyle\frac{1}{2}.\]

Instead we must recognise that the probability that a person buys a ticket to see a Sci-Fi film is the probability that a person buys a ticket to see either Forgotten or The Diamond Valley.

Therefore to calculate this probability, we add the probabilities of a person buying a ticket to see each of these films:

\[
\begin{align}
P(\text{Sci-Fi}) &= P(\text{Forgotten Game})+P(\text{The Diamond Valley})\\
&= \var{Avatar}+\var{SW}\\
&= \var{Avatar+SW}.
\end{align}
\]

Then the expected number of people who bought tickets to see a Sci-Fi film is 

\[
\var{Avatar+SW} \times \var{no_people} = \var{({Avatar+SW})*{no_people}}.
\]