Measures of dispersion  (16.416.6)
Understand the concept of standard deviation for both grouped and ungrouped data sets

Theory

Examples(a) Find the standard deviation of each of the following sets of data.
(Give your answers correct to 3 significant figures.)
(i) 3, 5, 7, 9, 11, 13
(ii) 1, 2, 2, 3, 5
(b) A fitness centre records the number of visits to the centre from
some members last week. The following shows the number of visits from
two groups of members aged 18  30 and 31  40.
Members aged 18  30
Number of visits Frequency 1 13 2 40 3 20 4 48 5 39 Number of visits Frequency 1 38 2 48 3 51 4 18 5 5
each group of members. (Give your answers correct to 3 sig. fig. if necessary.)
(ii) Which group of members have a more consistent number of visits to
the centre? Explain briefly.

Solutions(a)(i) Mean
\(\begin{array}{l} = \frac{{3 + 5 + 7 + 9 + 11 + 13}}{6}\\ = 8\end{array}\)
Standard deviation
\(\begin{array}{l} = \sqrt {\frac{{{{(3  8)}^2} + {{(5  8)}^2} + {{(7  8)}^2} + {{(9  8)}^2} + {{(11  8)}^2} + {{(13  8)}^2}}}{6}} \\ = \sqrt {\frac{{25 + 9 + 1 + 1 + 9 + 25}}{6}} \\ = \sqrt {\frac{{70}}{6}} \end{array}\)
\( = \underline{\underline {3.42}} \) (corr. to 3 sig. fig.)
(ii) Mean
\(\begin{array}{l} = \frac{{1 + 2 + 2 + 3 + 5}}{5}\\ = 2.6\end{array}\)
Standard deviation
\(\begin{array}{l} = \sqrt {\frac{{{{(1  2.6)}^2} + {{(2  2.6)}^2} + {{(2  2.6)}^2} + {{(3  2.6)}^2} + {{(5  2.6)}^2}}}{5}} \\ = {\sqrt {\frac{{2.56 + 0.36 + 0.36 + 0.16 + 5.76}}{5}} ^{}}\\ = {\sqrt {\frac{{9.2}}{5}} ^{}}\end{array}\)
\( = \underline{\underline {1.36}} \) (corr. to 3 sig. fig.)
(b)(i) For members aged 18  30,
mean\( = \underline{\underline {3.375}} \)
standard deviation\( = \underline{\underline {1.31}} \) (corr. to 3 sig. fig.)
For members aged 31  40,
mean\( = \underline{\underline {2.4}} \)
standard deviation\( = \underline{\underline {1.06}} \) (corr. to 3 sig. fig.)
(ii)∵ standard deviation of the number of visits from members aged 31  40
is smaller than that from members aged 18  30
∴members aged 31  40 have a more consistent number of visits to the centre.
Compare the dispersions of different sets of data using appropriate measures

Theory

Examples

Solutions
Understand the applications of standard deviation to reallife problems involving standard scores and the normal distribution

Theory

Examples(a) The following table shows the means and standard deviations of the time
taken by Tony to complete swimming 100 m and 400 m.
Swimming event Mean Standard deviation 100 m 4.35 min 0.45 min 400 m 17.5 min 1.5 min
400 m were 4.7 min and 18.1 min respectively. In which swimming event
did Tony perform better yesterday?
(b)The lengths of a batch of strings are normally distributed with a
mean of 50 mm and a standard deviation of 2 mm. Find the percentage
of strings with lengths in each of the following ranges.
(i) Between 48 mm and 50 mm
(ii) Less than 46 mm

Solutions(a) Standard score of the time taken by Tony to complete swimming 100 m \( = \frac{{4.7  4.35}}{{0.45}}\)
\( = 0.778\) (corr. to 3 sig. fig.)
Standard score of the time taken by Tony to complete swimming 400 m
\(\begin{array}{l} = \frac{{18.1  17.5}}{{1.5}}\\ = {0.4^{}}\end{array}\)
∵Standard score for the time taken to complete swimming 400 m
is greater than standard score for the time taken to complete swimming 100 m
∴ Tony performed better in the swimming event of 400 m.
(b)(i)It is given that \(\bar x = 50{\rm{ mm}}\) and \(\sigma = 2{\rm{ mm}}\).
\(\because \bar x  \sigma = 48{\rm{ mm}}\)
∴Percentage of strings with lengths between 48 mm and 50 mm
=Percentage of data between \(\bar x  \sigma \) and \(\bar x\)
\( = \underline{\underline {34\% }} \)
(ii)\(\because \bar x  2\sigma = 46{\rm{ mm}}\)
∴Percentage of strings with lengths less than 46 mm
=Percentage of data below \(\bar x  2\sigma \)
\(\begin{array}{l} = 50\%  34\%  13.5\% \\ = {\underline{\underline {2.5\% }} ^{}}\end{array}\)