Question

$100$ surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surname was obtained as follows:

Number of letters	$1 - 4$	$4 - 7$	$7 - 10$	$10 - 13$	$13 - 16$	$16 - 19$
Number of surnames	$6$	$30$	$40$	$16$	$4$	$4$

Determine the median number of letters in the surnames. Find the mean number of letters in the surname? Also, find the modal size of the surnames.

Answer 1

We will use the formula of median to find the median of the given information. The formula used to find the median of a given data is as follows: $Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
Where $l$ is the lower limit of median class, $n$ is the sum of all frequencies, $cf$ is the cumulative frequency before the median class, $f$ is the frequency of median class and $h$ is the size of median class.

Complete step-by-step answer:
(1)Calculation of median: The class intervals with respective cumulative frequencies can be represented as follows:-

Number of letters	Frequency $\left( f \right)$	Cumulative frequency $\left( {cf} \right)$
$1 - 4$	$6$	$6$
$4 - 7$	$30$	$36$
$7 - 10$	$40$	$76$
$10 - 13$	$16$	$92$
$13 - 16$	$4$	$96$
$16 - 19$	$4$	$100$
$n = \sum {f = 100}$

From the table, we obtain $n = 100 \Rightarrow \dfrac{n}{2} = 50$
Cumulative frequency $\left( {cf} \right)$just greater than $\dfrac{n}{2}$ $\left( {i.e.,50} \right)$ is $76$, which lies in the interval $7 - 10$.
Therefore, median class=$7 - 10$
Lower limit of the median class, $l = 7$
Frequency of the median class, $f = 40$
Cumulative frequency of the class preceding the median class, $cf = 36$
Class size, $h = 3$
Therefore, $Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
$Median = 7 + \left( {\dfrac{{50 - 36}}{{40}}} \right) \times 3$
$\Rightarrow Median = 7 + \dfrac{{14 \times 3}}{{40}}$
$\Rightarrow Median = 7 + \dfrac{{42}}{{40}}$
$\Rightarrow Median = 8.05$
Therefore, the median number of letters in the surnames is $8.05$.
(2) Calculation of mean: To calculate class marks of the given class intervals, the following relation is used:
$x = \dfrac{{Upper{\text{ }}limit{\text{ }} + {\text{ }}Lower{\text{ }}limit}}{2}$

Number of letters	Frequency $\left( f \right)$	$x$	$f \times x$
$1 - 4$	$6$	$2.5$	$15$
$4 - 7$	$30$	$5.5$	$165$
$7 - 10$	$40$	$8.5$	$340$
$10 - 13$	$16$	$11.5$	$184$
$13 - 16$	$4$	$14.5$	$58$
$16 - 19$	$4$	$17.5$	$70$
$n = \sum {f = 100}$		$\sum {fx = 832}$

From the table, we obtain
$n = \sum {f = 100}$ and $\sum {fx = 832}$
Therefore, $Mean = \dfrac{{\sum {fx} }}{{\sum f }}$
$Mean = \dfrac{{832}}{{100}}$
$\Rightarrow Mean = 8.32$
Therefore, the mean number of letters in the surname is $8.32$.
(3) Calculation of modal size: The data given can be written as:

Number of letters	Frequency $\left( f \right)$
$1 - 4$	$6$
$4 - 7$	$30$
$7 - 10$	$40$
$10 - 13$	$16$
$13 - 16$	$4$
$16 - 19$	$4$
$n = \sum {f = 100}$

From the table, it can be observed that the maximum class frequency is $40$, which lies in the interval $7 - 10$.
Therefore, modal class=$7 - 10$
Lower limit of the modal class $l = 7$
Frequency of the modal class, ${f_1} = 40$
Frequency of the class preceding the modal class, ${f_0} = 30$
Frequency of the class succeeding the modal class, ${f_2} = 16$
Class size, $h = 3$
Therefore, $Mode + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h$
$Mode = 7 + \left( {\dfrac{{40 - 30}}{{2\left( {40} \right) - 30 - 16}}} \right) \times 3$
$\Rightarrow$ $Mode = 7 + \dfrac{{10}}{{34}} \times 3$
$\Rightarrow Mode = 7 + \dfrac{{30}}{{34}}$
$\Rightarrow Mode = 7.88$

Therefore, the modal size of the surnames is $7.88$.

Note: The class with maximum frequency is called the modal class. The mode is a value inside the modal class and is given by the formula:
$Mode = l + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h$
Where $l$ is the lower limit of modal class, ${f_1}$ is the frequency of the modal class, ${f_0}$ is the frequency of class preceding the modal class, ${f_2}$ is the frequency of class succeeding the modal class and $h$ is the class size.