a) Order the number from smallest to largest:

\(\displaystyle{6},\ {7},{8},{8},\ {9},\ {9},\ {9},\ {9},\ {9},\ {9},\ {9},\ {10},\ {10},\ {10},\ {10},\ {10},\ {10},\ {10},\ {10},\ {11},\ {11},\ {11},\ {11},\ {11},\ {11},\ {11},\ {11},\ {11},\ {12},\ {12},\ {12},\ {12},\ {12},\ {13},\ {13},\ {13},\ {13},\ {13},\ {14},\ {14},\ {14},\ {14},\ {14},\ {15},\ {15},\ {15},\ {15},\ {16},\ {16},\ {16}.\)

Since the number of scores is even, the median is the average of the middle scores:

\(\displaystyle{M}={Q}_{{{2}}}={\frac{{{11}\ +\ {11}}}{{{2}}}}={11}\)

The first quartile is the median of the data values below the median (or at \(\displaystyle{25}\%\) of the data):

\(\displaystyle{Q}_{{{1}}}={10}\)

The third quartile is the median of the data values above the median (or at \(\displaystyle{75}\%\) of the data):

\(\displaystyle{Q}_{{{3}}}={13}\)

The interquartile range IQR is the difference of the third and first quartile:

\(\displaystyle{I}{Q}{R}={13}\ -\ {10}={3}\)

The whiskers of the boxplot are at the minimum and maximum value. The box starts at the lower quartile, end at the upper quartile and has a vertical line at the median.

The lower quartile is at \(\displaystyle{25}\%\) of the sorted data list.

The median at \(\displaystyle{50}\%\) and upper quartile:

at \(\displaystyle{75}\%\)

b) The class width is the difference between the largest and smallest value, divided by the number of classes (round up to the nearest integer!).

\(\displaystyle{C}{l}{a}{s}{s}\ {W}{i}{\left.{d}{t}\right.}{h}={\frac{{{16}\ -\ {6}}}{{{4}}}}={2.5}\ \approx\ {3}\)

Determine the midpoints, frequencies, the products of the midpoints and the frequaencies, and the products of the squared midpoints and the frequencies.

\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Interval}&\text{Midpoint x}&{f}&{x}{f}&{x}^{{{2}}}{f}\backslash{h}{l}\in{e}{6}-{8}&{7}&{4}&{28}&{196}\backslash{h}{l}\in{e}{9}-{11}&{10}&{24}&{240}&{2400}\backslash{h}{l}\in{e}{12}-{14}&{13}&{15}&{195}&{2535}\backslash{h}{l}\in{e}{15}-{17}&{16}&{7}&{112}&{1792}\backslash{h}{l}\in{e}&&&&\backslash{h}{l}\in{e}&{S}{U}{M}&{50}&{575}&{6923}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\)

The sample mean is then:

\(\displaystyle\overline{{{x}}}={\frac{{\sum\ {x}{f}}}{{{n}}}}={\frac{{{575}}}{{{50}}}}={11.5}\)

The sample standard deviation is then:

\(\displaystyle{s}=\sqrt{{{\frac{{\sum\ {x}^{{{2}}}{f}\ -\ \frac{{\left(\sum\ {x}{f}\right)}^{{{2}}}}{{n}}}}{{{n}\ -\ {1}}}}}}=\sqrt{{{\frac{{{6923}\ -\ \frac{{\left({575}\right)}^{{{2}}}}{{50}}}}{{{50}\ -\ {1}}}}}}\ \approx\ {2.5173}\)

Using Chebyshev's Rule with \(\displaystyle{k}={2}\), we know that at least

\(\displaystyle{100}{\left({1}\ -\ {\frac{{{1}}}{{{k}^{{{2}}}}}}\right)}\%={100}{\left({1}\ -\ {\frac{{{1}}}{{{4}}}}\right)}\%={75}\%\)

is within 2 standard deviations from the mean.

\(\displaystyle\overline{{{x}}}\ -\ {2}{s}={11.5}\ -\ {2}{\left({2.5173}\right)}={6.4654}\)

\(\displaystyle\overline{{{x}}}\ +\ {2}{s}={11.5}\ +\ {2}{\left({2.5173}\right)}={16.5346}\)

c) n is the number of values in the data set.

\(\displaystyle{n}={50}\)

The mean is the sum of all values divided by the number of values:

\(\displaystyle\overline{{{x}}}={\frac{{{15}\ +\ {14}\ +\ {14}\ +\ \cdots\ +\ {10}\ +\ {11}\ +\ {9}}}{{{50}}}}={11.48}\)

The variance is the sum of squared deviations from the mean divided by \(\displaystyle{n}\ -\ {1}.\) The standard deviation is the square root of the variance:

\(\displaystyle{s}=\sqrt{{{\frac{{{\left({15}\ -\ {11.48}\right)}^{{{2}}}\ +\ \cdots\ +\ {\left({9}\ -\ {11.48}\right)}^{{{2}}}}}{{{50}\ -\ {1}}}}}}\ \approx\ {2.4431}\)

\(\displaystyle{6},\ {7},{8},{8},\ {9},\ {9},\ {9},\ {9},\ {9},\ {9},\ {9},\ {10},\ {10},\ {10},\ {10},\ {10},\ {10},\ {10},\ {10},\ {11},\ {11},\ {11},\ {11},\ {11},\ {11},\ {11},\ {11},\ {11},\ {12},\ {12},\ {12},\ {12},\ {12},\ {13},\ {13},\ {13},\ {13},\ {13},\ {14},\ {14},\ {14},\ {14},\ {14},\ {15},\ {15},\ {15},\ {15},\ {16},\ {16},\ {16}.\)

Since the number of scores is even, the median is the average of the middle scores:

\(\displaystyle{M}={Q}_{{{2}}}={\frac{{{11}\ +\ {11}}}{{{2}}}}={11}\)

The first quartile is the median of the data values below the median (or at \(\displaystyle{25}\%\) of the data):

\(\displaystyle{Q}_{{{1}}}={10}\)

The third quartile is the median of the data values above the median (or at \(\displaystyle{75}\%\) of the data):

\(\displaystyle{Q}_{{{3}}}={13}\)

The interquartile range IQR is the difference of the third and first quartile:

\(\displaystyle{I}{Q}{R}={13}\ -\ {10}={3}\)

The whiskers of the boxplot are at the minimum and maximum value. The box starts at the lower quartile, end at the upper quartile and has a vertical line at the median.

The lower quartile is at \(\displaystyle{25}\%\) of the sorted data list.

The median at \(\displaystyle{50}\%\) and upper quartile:

at \(\displaystyle{75}\%\)

b) The class width is the difference between the largest and smallest value, divided by the number of classes (round up to the nearest integer!).

\(\displaystyle{C}{l}{a}{s}{s}\ {W}{i}{\left.{d}{t}\right.}{h}={\frac{{{16}\ -\ {6}}}{{{4}}}}={2.5}\ \approx\ {3}\)

Determine the midpoints, frequencies, the products of the midpoints and the frequaencies, and the products of the squared midpoints and the frequencies.

\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Interval}&\text{Midpoint x}&{f}&{x}{f}&{x}^{{{2}}}{f}\backslash{h}{l}\in{e}{6}-{8}&{7}&{4}&{28}&{196}\backslash{h}{l}\in{e}{9}-{11}&{10}&{24}&{240}&{2400}\backslash{h}{l}\in{e}{12}-{14}&{13}&{15}&{195}&{2535}\backslash{h}{l}\in{e}{15}-{17}&{16}&{7}&{112}&{1792}\backslash{h}{l}\in{e}&&&&\backslash{h}{l}\in{e}&{S}{U}{M}&{50}&{575}&{6923}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\)

The sample mean is then:

\(\displaystyle\overline{{{x}}}={\frac{{\sum\ {x}{f}}}{{{n}}}}={\frac{{{575}}}{{{50}}}}={11.5}\)

The sample standard deviation is then:

\(\displaystyle{s}=\sqrt{{{\frac{{\sum\ {x}^{{{2}}}{f}\ -\ \frac{{\left(\sum\ {x}{f}\right)}^{{{2}}}}{{n}}}}{{{n}\ -\ {1}}}}}}=\sqrt{{{\frac{{{6923}\ -\ \frac{{\left({575}\right)}^{{{2}}}}{{50}}}}{{{50}\ -\ {1}}}}}}\ \approx\ {2.5173}\)

Using Chebyshev's Rule with \(\displaystyle{k}={2}\), we know that at least

\(\displaystyle{100}{\left({1}\ -\ {\frac{{{1}}}{{{k}^{{{2}}}}}}\right)}\%={100}{\left({1}\ -\ {\frac{{{1}}}{{{4}}}}\right)}\%={75}\%\)

is within 2 standard deviations from the mean.

\(\displaystyle\overline{{{x}}}\ -\ {2}{s}={11.5}\ -\ {2}{\left({2.5173}\right)}={6.4654}\)

\(\displaystyle\overline{{{x}}}\ +\ {2}{s}={11.5}\ +\ {2}{\left({2.5173}\right)}={16.5346}\)

c) n is the number of values in the data set.

\(\displaystyle{n}={50}\)

The mean is the sum of all values divided by the number of values:

\(\displaystyle\overline{{{x}}}={\frac{{{15}\ +\ {14}\ +\ {14}\ +\ \cdots\ +\ {10}\ +\ {11}\ +\ {9}}}{{{50}}}}={11.48}\)

The variance is the sum of squared deviations from the mean divided by \(\displaystyle{n}\ -\ {1}.\) The standard deviation is the square root of the variance:

\(\displaystyle{s}=\sqrt{{{\frac{{{\left({15}\ -\ {11.48}\right)}^{{{2}}}\ +\ \cdots\ +\ {\left({9}\ -\ {11.48}\right)}^{{{2}}}}}{{{50}\ -\ {1}}}}}}\ \approx\ {2.4431}\)