Math, asked by khushboogoyal0177, 2 months ago

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

Answered by kimtaehyung1730
0

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

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