
$100$ 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.
| Number of letters | $1 - 4$ | $4 - 7$ | $7 - 10$ | $10 - 13$ | $13 - 16$ | $16 - 19$ |
| Number of surnames | $6$ | $30$ | $40$ | $16$ | $4$ | $4$ |
Answer
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Hint: 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:-
From the table, we obtain $n = 100 \Rightarrow \dfrac{n}{2} = 50$
Cumulative frequency $\left( {cf} \right)$just greater than $\dfrac{n}{2}$$\left( {i.e.,50} \right)$ is $76$, which lies in the interval $7 - 10$.
Therefore, median class=$7 - 10$
Lower limit of the median class, $l = 7$
Frequency of the median class, $f = 40$
Cumulative frequency of the class preceding the median class, $cf = 36$
Class size, $h = 3$
Therefore, $Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
$Median = 7 + \left( {\dfrac{{50 - 36}}{{40}}} \right) \times 3$
$ \Rightarrow Median = 7 + \dfrac{{14 \times 3}}{{40}}$
$ \Rightarrow Median = 7 + \dfrac{{42}}{{40}}$
$ \Rightarrow Median = 8.05$
Therefore, the median number of letters in the surnames is $8.05$.
(2) Calculation of mean: To calculate class marks of the given class intervals, the following relation is used:
$x = \dfrac{{Upper{\text{ }}limit{\text{ }} + {\text{ }}Lower{\text{ }}limit}}{2}$
From the table, we obtain
$n = \sum {f = 100} $ and $\sum {fx = 832} $
Therefore, $Mean = \dfrac{{\sum {fx} }}{{\sum f }}$
$Mean = \dfrac{{832}}{{100}}$
$ \Rightarrow Mean = 8.32$
Therefore, the mean number of letters in the surname is $8.32$.
(3) Calculation of modal size: The data given can be written as:
From the table, it can be observed that the maximum class frequency is $40$, which lies in the interval $7 - 10$.
Therefore, modal class=$7 - 10$
Lower limit of the modal class $l = 7$
Frequency of the modal class, ${f_1} = 40$
Frequency of the class preceding the modal class, ${f_0} = 30$
Frequency of the class succeeding the modal class, ${f_2} = 16$
Class size, $h = 3$
Therefore, $Mode + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h$
$Mode = 7 + \left( {\dfrac{{40 - 30}}{{2\left( {40} \right) - 30 - 16}}} \right) \times 3$
$ \Rightarrow $$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.
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.
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