Solution. \[\overline {X}=\frac {1}{n}\sum _{i=1}^{5}X_i=\frac {0+5+2+3+5}{5}=\frac {15}{5}=3.\]

Solution. Add an \(fX\) row and total both:

\[\overline {X}=\frac {\sum fX}{\sum f}=\frac {9}{8}=1.125.\]

Solution. The mid-point of a class is the average of its two limits: for \(1\)–\(5\) that is \((1+5)/2=3\), and so on.

\[\overline {X}=\frac {\sum fX}{\sum f}=\frac {71}{7}=10.14.\]

Solution. The three parts are the same calculation at three levels of summary, and each level costs a little accuracy.

(a). Raw values. Every observation is present, so the mean is exact: \[\overline {X}=\frac {1}{n}\sum X=\frac {15+20+21+20+36+15+25+15}{8}=\frac {167}{8}=20.875.\]

(b). Frequency table. The individual values are still known; they have only been counted rather than listed. Multiply each value by how often it occurs:

\[\overline {X}=\frac {\sum fX}{\sum f}=\frac {26}{15}=1.733.\] This is still exact – writing \(2\) three times or writing “\(2\), three times” is the same information.

(c). Grouped data. Here the individual values are gone. All we know is how many fell in each class, so each observation is taken to sit at its class mid-point:

\[\overline {X}=\frac {\sum fX}{\sum f}=\frac {299}{18}=16.61.\]

This last answer is an estimate, not the mean of the original data. Treating every observation in \(11\)–\(15\) as though it were exactly \(13\) is only right on average, and the error does not necessarily cancel. Once data have been grouped, the exact mean cannot be recovered – which is worth remembering before grouping data you still need to work with.

Solution. (a) Order the values first. Nothing about a median means anything until you do. \[15,\; 15,\; 15,\; 20,\; 20,\; 21,\; 25,\; 36\] There are \(n=8\) observations. Since \(n\) is even the median is not one of them; it is the average of the \(4^{\text {th}}\) and \(5^{\text {th}}\) values: \[\text {Median}=\frac {20+20}{2}=20\]

(b) Add a cumulative frequency column.

Here \(n=\sum f=15\), which is odd, so the median is one of the observations: the \(\tfrac {1}{2}(n+1)=8^{\text {th}}\) value. Reading down the \(Cf\) column, the count first reaches \(8\) at \(X=1\), so \[\text {Median}=1\]

(c) The individual values are gone, so the median can no longer be read off a list — it has to be located inside a class.

Here \(n=18\), so \(\tfrac {n}{2}=9\). The class median is the first class whose cumulative frequency reaches \(9\), which is \(11-15\). For that class \[L_m=10.5,\hspace {0.8cm} F=4,\hspace {0.8cm} f_m=5,\hspace {0.8cm} i=5\] and the formula set out in the next subsection gives \[\text {Median}=L_m+\Bigg (\frac {\frac {n}{2}-F}{f_m}\Bigg )i =10.5+\Bigg (\frac {9-4}{5}\Bigg )5=15.5\]

The answer lands exactly on the upper boundary of the class, which looks odd but is correct here: the cumulative frequency reaches \(\tfrac {n}{2}=9\) precisely at the end of the class \(11-15\), so the interpolation runs the whole way across it. Notice also that the classes are written \(1-5\), \(6-10\) and so on, yet the boundaries used are \(0.5\), \(5.5\), \(10.5\). The gap between \(5\) and \(6\) is an artefact of recording whole numbers; the underlying scale is continuous, and it is the boundaries that the formula needs.

Solution. Step 1: Construct the cumulative frequency distribution.

\(\displaystyle {\frac {n}{2}=\frac {50}{2}=25}\implies \) Class median is the \(3^{\text {rd}}\) class.

So \(F=22,\hspace {0.2cm} f_m=12,\hspace {0.2cm} L_m=21.5\hspace {0.2cm}\) and \(\hspace {0.2cm}i=10\). Therefore

\begin {align*} \text {Median} & = L_m + \Bigg (\frac {\frac {n}{2}-F}{f_m}\Bigg )\, i\\\\ & = 21.5 + \Bigg (\frac {25-22}{12}\Bigg )\times 10\\ & = 24 \end {align*}

Thus, 25 persons take less than 24 minutes to travel to work and another 25 persons take more than 24 minutes to travel to work.

Solution. Step 1: Construct the cumulative frequency distribution.

Step 2: Determine the \(Q_1\) and \(Q_3\)

Class \(\displaystyle {Q_1 = \frac {n}{4} = \frac {50}{4}=12.5}\implies \) Class \(Q_1\) is the \(2^{\text {nd}}\) class. Therefore, \begin {align*} Q_1 & = L_{Q_1} + \Bigg (\frac {\frac {n}{4}-F}{f_{Q_1}}\Bigg )\, i\\\\ & = 10.5 + \Bigg (\frac {12.5-8}{14}\Bigg )\times 10\\\\ \implies Q_1 & = 13.7143\\ \end {align*}

Solution. (a). Every value occurs once, so no value occurs more often than the others and there is no mode. This is worth noticing: unlike the mean and the median, the mode need not exist.

(b). Both \(0\) and \(1\) occur three times, and nothing occurs more often, so there are two modes – \(0\) and \(1\). Such a distribution is called bimodal. Here too the mode differs from the other averages: it need not be unique.

(c). The largest frequency is \(7\), against \(X=0\), so the mode is 0.

(d). The data are grouped, so what can be identified is a modal class rather than a single modal value — the individual observations are no longer visible. The largest frequency is \(3\), and three classes share it: \[\textbf {modal classes: } 6-10,\ 11-15\ \text {and}\ 16-20.\] With three of the four classes tied, the mode tells us almost nothing here. That is the honest conclusion: an average that cannot distinguish three quarters of the data is the wrong summary for it, and the median or mean would serve better.

Solution. Based on the table \(L_{m_0} = 10.5\hspace {0.1cm},\hspace {0.2cm} \Delta _1 = (14-8)=6\hspace {0.1cm},\hspace {0.2cm}\Delta _2 = (14-12)=2\hspace {0.2cm}\) and \(\hspace {0.2cm} i = 10\)

\begin {align*} \text {Mode} & = 10.5 + \Bigg (\frac {6}{6 + 2}\Bigg )\times 10\\ & = 17.5\\ \end {align*}