Question

If the median of the distribution given below is 28.5. Find the value of x and y. ​

Answer 1

Answer:

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Answer 2

Appropriate Question

The mean of the following data is 28.5 and sum of the frequencies is 60.

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c}\sf Class\: interval&\sf Frequency\: (f)\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf 0 - 10&\sf 5\\\\\sf 10 - 20 &\sf x\\\\\sf 20-30 &\sf 20\\\\\sf 30 - 40&\sf 15\\\\\sf 40-50&\sf y\\\\\sf 50-60&\sf 5\\\frac{\qquad}{}&\frac{\qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

Find the value of x and y.

$\large\underline{\sf{Solution-}}$

The given frequency distribution table is

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf Class\: interval&\sf Frequency\: (f)&\sf \: cumulative \: frequency\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf 0 - 10&\sf 5&\sf5\\\\\sf 10 - 20 &\sf x&\sf5 + x\\\\\sf 20-30 &\sf 20&\sf25 + x\\\\\sf 30 - 40&\sf 15&\sf40 + x\\\\\sf 40-50&\sf y&\sf40 + x + y\\\\\sf 50-60&\sf 5&\sf45 + x + y\\\frac{\qquad}{}&\frac{\qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

Now, given that,

Sum of all the frequency is 60.

$\rm :\longmapsto\: \sum \: f = 60$

$\rm :\longmapsto\:45 + x + y = 60$

$\rm :\longmapsto\: x + y = 60 - 45$

$\rm :\longmapsto\: x + y = 15 - - - (1)$

We know,

Median formula is

$\boxed{ \bf M= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}}$

Here,

• l denotes lower limit of median class

• h denotes width of median class

• f denotes frequency of median class

• cf denotes cumulative frequency of the class preceding the median class

• N denotes sum of frequency

According to the given data,

Median = 28.5

Median class is 20-30

So,

l = 20,

h = 10,

f = 20,

cf = cf of preceding class = 5 + x

and

N/2 = 30

By substituting all the given values in the formula,

$\dashrightarrow\sf M= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}$

$\dashrightarrow\sf 28.5= 20 + \Bigg \{10 \times \dfrac{ \bigg( 30 - (5 + x) \bigg)}{20} \Bigg \}$

$\dashrightarrow\sf 28.5 - 20 = \Bigg \{ \dfrac{ \bigg( 30 - 5 - x \bigg)}{2} \Bigg \}$

$\dashrightarrow\sf 8.5 = \Bigg \{ \dfrac{ \bigg( 25 - x \bigg)}{2} \Bigg \}$

$\rm :\longmapsto\:17 = 25 - x$

$\rm :\longmapsto\:x = 25 - 17$

$\bf :\implies\:x = 8$

On substituting the value of x, in equation (1), we get

$\rm :\longmapsto\:8 + y = 15$

$\bf\implies \:y = 15 - 8 = 7$

Additional Information :-

Mean using Direct Method :-

$\dashrightarrow\sf Mean = \dfrac{ \sum f_i x_i}{ \sum f_i}$

Mean using Short Cut Method :-

$\dashrightarrow\sf Mean =A \: + \: \dfrac{ \sum f_i d_i}{ \sum f_i}$

Mean using Step Deviation Method :-

$\dashrightarrow\sf Mean =A \: + \: \dfrac{ \sum f_i u_i}{ \sum f_i} \times h$