Question

An article in Technometrics (Vol. 19, 1977, p. 425) presents the following data on the motor fuel octane ratings of several blends of gasoline:
88.5 98.8 89.6 92.2 92.7 91.8 91.0 91.0 94.7 88.3 90.4 83.4 87.9 88.4 87.5 90.9 84.3
90.4 91.6 91.0 93.0 92.6 87.8 89.9 90.1 91.2 90.7 88.2 94.4 93.7 88.3 91.8 89.0 90.6
88.6 88.5 90.4 96.5 89.2 89.7 89.8 92.2 88.3 93.3 91.2 84.3 92.3 92.2 91.6 87.7 94.2
87.4 91.2 93.2 88.9 90.3 91.1 85.3 91.1 86.7 88.6 89.8 90.0 86.7 90.1 90.5 94.2 88.7
92.7 91.5 93.4 89.3 100.3 90.8 92.7 93.3 89.9 96.1 91.1 87.6 90.1 89.3 86.7

Required:
Construct a stem-and-leaf display for these data. Calculate the median and quartiles of these data.

Answer 1

Answer:

$Median = 90.4$

$Q_1 = 88.6$

$Q_3 = 92.2$

Step-by-step explanation:

Given

The above data

Required

- A stem and leaf display

- The median

- The quartiles

First, determine the range of the data

$Smallest = 83.4$

$Highest = 100.3$

Next, group each dataset base on common whole numbers.

So, we have:

$83.4$

$84.3\ 84.3$

$85.3$

$86.7\ 86.7\ 86.7$

$87.4\ 87.5\ 87.6\ 87.7\ 87.8\ 87.9$

$88.2\ 88.3\ 88.3\ 88.3\ 88.4\ 88.5\ 88.5\ 88.6\ 88.6\ 88.7\ 88.9$

$89.0\ 89.2\ 89.3\ 89.3\ 89.6\ 89.7\ 89.8\ 89.8\ 89.9\ 89.9$

$90.0\ 90.1\ 90.1\ 90.1\ 90.3\ 90.4\ 90.4\ 90.4\ 90.5\ 90.6\ 90.7\ 90.8\ 90.9$

$91.0\ 91.0\ 91.0\ 91.1\ 91.1\ 91.1\ 91.2\ 91.2\ 91.2\ \ 91.5\ 91.6\ 91.6\ 91.8\ 91.8$

$92.2\ 92.2\ 92.2\ 92.3\ 92.6\ 92.7\ 92.7\ 92.7$

$93.0\ 93.2\ 93.3\ 93.3\ 93.4\ 93.7$

$94.2\ 94.2\ 94.4\ 94.7$

$96.1\ 96.5$

$98.8$

$100.3$

Next, we construct the stem and leaf plot.

The whole numbers will be the stem while the decimal parts will be the leaf.

So, we have:

$\begin{array}{ccc}{Stem} & {} & {Leaf} & {83} & {|} & {.4} & {84} & {|} & {.3\ .3} & {85} & {|} & {.3} & {86} & {|} & {.7\ .7\ .7} & {87} &{|} & {.4\ .5\ .6\ .7\ .8\ .9} & {88} & {|} & {.2\ .3\ .3\ .3\ .4\ .5\ .5\ .6\ .6\ .7\ .9} &{89} & {|} & {.0\ .2\ .3\ .3\ .6\ .7\ .8\ .8\ .9\ .9} & {90} & {|} &{.0\ .1\ .1\ .1\ .3\ .4\ .4\ .4\ .5\ .6\ .7\ .8\ .9} & {91} &{|}&{.0\ .0\ .0\ .1\ .1\ .1\ .2\ .2\ .2\ .5\ .6\ .6\ .8\ .8} &{92} &{|} &{.2\ .2\ .2\ .3\ .6\ .7\ .7\ .7} \ \end{array}$

  $\begin{array}{ccc} {93} & {|} & {.0\ .2\ .3\ .3\ .4\ .7} & {94} &{|} & {.2\ .2\ .4\ .7} &{96} & {|} & {.1\ .5} & {98} & {|} & {.8} & {100} &{|} &{.3} \ \end{array}$

From the above plot,

$n = 83$

The median is calculated as:

$Median = \frac{n+1}{2}th$

$Median = \frac{83+1}{2}th$

$Median = \frac{84}{2}th$

$Median = 42nd$

i.e. the median is at the 42nd position.

From the above stem and leaf plot.

The 42nd position is at stem 90 and the leaf  .4

So the median is:

$Median = 90.4$

The lower quartile (Q) is calculated as:

$Q_1 = \frac{n+1}{4}th$

$Q_1 = \frac{83+1}{4}th$

$Q_1 = \frac{84}{4}th$

$Q_1 = 21st$

i.e. the lower quartile is at the 21st position.

From the above stem and leaf plot.

The 42nd position is at stem 88 and the leaf .6

So the lower quartile is:

$Q_1 = 88.6$

The upper quartile (Q3) is calculated as:

$Q_3 = 3 * \frac{n+1}{4}th$

$Q_3 = 3 * \frac{83+1}{4}th$

$Q_3 = 3 * \frac{84}{4}th$

$Q_3 = 3 * 21th$

$Q_3 = 63rd$

i.e. the upper quartile is at the 63rd position.

From the above stem and leaf plot.

The 63rd position is at stem 92 and the leaf .2

So the upper quartile is:

$Q_3 = 92.2$