are
of the following distribution Median I
values
33.5 and 34
respectively find missing frequentes
and verify that the results
Values 0-10 10-20 20-30 30-40 40-50 50-60 69-70
Frequency 4
16
F1
F2
f3
6
6
Answers
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,
\begin{gathered}30 + x + y + z = 230 \\ \\ x + y + z = 200 \: \: \: ....eq1 \\ \\ \end{gathered}
30+x+y+z=230
x+y+z=200....eq1
Mode=34
\begin{gathered}\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 \\ \\ \end{gathered}
Mode=l+(
2f
1
−f
0
−f
2
f
1
−f
0
)×h
34=30+(
2y−x−z
y−x
)×10
4=(
2y−x−z
y−x
)×10
8y−4x−4z=10y−10x
6x−2y−4z=0
3x−y−2z=0...eq2
Median=33.5
\begin{gathered}\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 \\ \\ \end{gathered}
Median=l+(
f
2
n
−cf
)×h
33.5=30+(
y
115−20−x
)×10
3.5=
y
95−x
×10
100
35y
=95−x
35y+100x=9500...eq3
from eq1 and eq2
\begin{gathered}2x +2y + 2z = 400 \\ 3x - y - 2z = 0 \\ - - - - - - \\ 5x + y = 400 \: \: \: eq4\end{gathered}
2x+2y+2z=400
3x−y−2z=0
−−−−−−
5x+y=400eq4
from eq3 and eq4
\begin{gathered} \\ (5x + y = 400) \times 20 \\ \\ 100x + 20y = 8000 \\ 100x + 35y = 9500\\ ( - ) \: \: \: \: ( - ) \: \: \: \: \: \: \: \: \: \: \: \: \: ( - ) \\ - - - - - - - \\ - 15y = - 1500 \\ \\ y = 100 \\ \\ \end{gathered}
(5x+y=400)×20
100x+20y=8000
100x+35y=9500
(−)(−)(−)
−−−−−−−
−15y=−1500
y=100
Now put value of y in eq3
\begin{gathered}5x + 100 = 400 \\ \\ 5x = 300 \\ \\ x = \frac{300}{5} \\ \\ x = 60 \\ \\ \end{gathered}
5x+100=400
5x=300
x=
5
300
x=60
put the value of x and y in eq1
\begin{gathered}60 + 100 + z = 200 \\ \\ z = 200 - 160 \\ \\ z = 40 \\ \end{gathered}
60+100+z=200
z=200−160
z=40