Question

14. The median and mode of the following wage distribution are known to be Rs. 33.5 and Rs. 34 respectively. Three frequency values from the table are however, missing. Find these missing values total 230.



Wages (in Rs)

0-10

10-20

20-30

30-40

40-50

50-60

60-70

Total



Frequency

4

16

?

?

?

6

4

230

Answer 1

Answer:

x=60

y=100

z=40

Step-by-step explanation:

Median and Modal class will be 30-40

Wages (in Rs)_____freq____CF

0-10________4___________4

10-20_______16________20

20-30_______x________20+x

30-40______y_________20+x+y

40-50______z________20+x+y+z

50-60______6_______26+x+y+z

60-70______4______30+x+y+z

Since total 230

So,

$30 + x + y + z = 230 \\ \\ x + y + z = 200 \: \: \: ....eq1$

Mode=34

$\boxed{\blue{\bold{Mode = l + \bigg( \frac{f_1 - f_0}{2f_1 - f_0 - f_2}\bigg ) \times h }}}\\ \\ 34 = 30 + \bigg( \frac{y -x }{2y - x - z} \bigg ) \times 10 \\ \\ 4 = ( \frac{y -x }{2y - x - z} ) \times 10 \\ \\ 8y- 4x- 4z = 10y - 10x \\ \\ 6x - 2y - 4z = 0 \\ \\ 3x - y - 2z = 0 \: \: \: ...eq2$

Median=33.5

$\boxed{\green{\bold{Median = l + \bigg(\frac{ \frac{n}{2} - cf }{f} \bigg) \times h}}} \\ \\ 33.5 = 30 + \bigg (\frac{115 -20 - x }{y}\bigg) \times 10 \\ \\ 3.5 = \frac{95 - x}{y} \times 10 \\ \\ \frac{35y}{100} = 95 - x \\ \\ 35y + 100x = 9500...eq3$

from eq1 and eq2

$2x +2y + 2z = 400 \\ 3x - y - 2z = 0 \\ - - - - - - \\ 5x + y = 400 \: \: \: eq4$

from eq3 and eq4

$\\ (5x + y = 400) \times 20 \\ \\ 100x + 20y = 8000 \\ 100x + 35y = 9500\\ ( - ) \: \: \: \: ( - ) \: \: \: \: \: \: \: \: \: \: \: \: \: ( - ) \\ - - - - - - - \\ - 15y = - 1500 \\ \\ y = 100$

Now put value of y in eq3

$5x + 100 = 400 \\ \\ 5x = 300 \\ \\ x = \frac{300}{5} \\ \\ x = 60$

put the value of x and y in eq1

$60 + 100 + z = 200 \\ \\ z = 200 - 160 \\ \\ z = 40$

Hope it helps you.

Answer 2

Concept

The median is that the middle value in a very list ordered from smallest to largest. The mode is that the most often occurring value on the list.

Given

The wage distribution is given where the median is $Rs.33.5$ and therefore the mode is $Rs.34$.

Find

We have to search out the missing frequencies.

Solution

Let the missing frequencies of the category intervals $20-30,30-40$ and $40-50$ be $f_{1},f_{2}$ and $f_{3}$ respectively.

Since $\sum f_{i}=230$.

Therefore,

$\begin{aligned}f_{1}+f_{2}+f_{3}&=\sum f_{i}-\left(4+16+6+4\right)\\ &=230-30\\ &=200\end{aligned}$                                    ......(1)

As the median is Rs. $33.5,$ so $30-40$ is that the median class.

$\begin{aligned}\text{Median }&=l+\frac{\frac{n}{2}+c.f.}{f}\times h\\ &=30+\frac{115-(4+16+f_{1})}{f_{2}}\times 10\\ 33.5 &=30+\frac{95+f_{1}}{f_{2}}\times 10\\ 10f_{1}+3.5f_{2}&=950\end{aligned}$                             ......(2)

Again mode is $34,$ therefore, $30-40$ is that the modal class and so

$\begin{aligned}\text{Mode }&=l+\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\times h\\ 34&=30+\frac{f_{2}-f_{1}}{2f_{2}-f_{1}-f_{3}}\times 10\\ 4&=\frac{10(f_{2}-f_{1})}{2f_{2}-f_{1}-f_{3}}\\ 3f_{1}-f_{2}-2f_{3}&=0\end{aligned}$                                 ......(3)

After solving equation (1), (2) and (3), we get

$f_{1}=60,f_{2}=100$ and $f_{3}=40$

Hence, the missing values is $60,100 \text{ and }40$.

#SPJ2