Explanation:
\large\underline{\sf{Solution-}}
Solution−
Calculations of Median
\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \small\boxed{\begin{array}{c |c|c} \tt{Classes} & \tt{Frequency}& \tt{c.f.} \\ \dfrac{\qquad\qquad}{ \sf 0-100} &\dfrac{\qquad\qquad}{ \sf 2} &\dfrac{\qquad\qquad}{ \sf 2}& \\ \dfrac{\qquad\qquad}{ \sf 100-200} &\dfrac{\qquad\qquad}{ \sf 5}&\dfrac{\qquad\qquad}{ \sf 7} & \\ \dfrac{\qquad\qquad}{ \sf 200-300} &\dfrac{\qquad\qquad}{ \sf x}&\dfrac{\qquad\qquad}{ \sf 7 + x} & \\ \dfrac{\qquad\qquad}{ \sf 300-400} &\dfrac{\qquad\qquad}{ \sf 12}&\dfrac{\qquad\qquad}{ \sf 19 + x} & \\ \dfrac{\qquad\qquad}{ \sf 400-500} &\dfrac{\qquad\qquad}{ \sf 17}&\dfrac{\qquad\qquad}{ \sf 36 + x} & \\ \dfrac{\qquad\qquad}{ \sf 500-600} &\dfrac{\qquad\qquad}{ \sf 20}&\dfrac{\qquad\qquad}{ \sf 56 + x} & \\ \dfrac{\qquad\qquad}{ \sf 600-700} &\dfrac{\qquad\qquad}{ \sf y}&\dfrac{\qquad\qquad}{ \sf 56 + x + y}\\ \dfrac{\qquad\qquad}{ \sf 700-800} &\dfrac{\qquad\qquad}{ \sf 9}&\dfrac{\qquad\qquad}{ \sf 65 + x + y}\\ \dfrac{\qquad\qquad}{ \sf 800-900} &\dfrac{\qquad\qquad}{ \sf 7}&\dfrac{\qquad\qquad}{ \sf 72 + x + y}\\ \dfrac{\qquad\qquad}{ \sf 900-1000} &\dfrac{\qquad\qquad}{ \sf 4}&\dfrac{\qquad\qquad}{ \sf 76 + x + y} &\end{array}} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \\ \end{gathered}
Classes
0−100
100−200
200−300
300−400
400−500
500−600
600−700
700−800
800−900
900−1000
Frequency
2
5
x
12
17
20
y
9
7
4
c.f.
7+x
19+x
36+x
56+x
56+x+y
65+x+y
72+x+y
76+x+y
We have given that,
\begin{gathered}\sf \: \sum \: f \: = \: 100 \\ \\ \end{gathered}
∑f=100
\begin{gathered}\sf \: 76 + x + y= \: 100 \\ \\ \end{gathered}
76+x+y=100
\begin{gathered}\sf \: x + y= \: 100 - 76\\ \\ \end{gathered}
x+y=100−76
\begin{gathered}\sf \: \sf \: \implies \: x + y= \: 24 - - - (1)\\ \\ \\ \end{gathered}
⟹x+y=24−−−(1)
Now, we know
\begin{gathered}\sf \: \sf \: \boxed{ \sf Median= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}} \\ \\ \end{gathered}
Median=l+{h×
f
(
N
−cf)
}
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
So, from above calculations, we have
Median = 525
Median class = 500 - 600
l = 500,
h = 100,
f = 20,
cf = cf of preceding class = 36 + x
N = 100
By substituting all the given values in the formula,
\begin{gathered}\sf \: Median= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \} \\ \\ \end{gathered}
\begin{gathered}\sf \: 525= 500 + \Bigg \{100\times \dfrac{ \bigg( 50 - (36 - x)\bigg)}{20} \Bigg \} \\ \\ \end{gathered}
525=500+{100×
(50−(36−x))
\begin{gathered}\sf \: 525 - 500 = 5\times (50 - 36 - x) \\ \\ \end{gathered}
525−500=5×(50−36−x)
\begin{gathered}\sf \: 25 = 5\times (14 - x) \\ \\ \end{gathered}
25=5×(14−x)
\begin{gathered}\sf \: 5 = 14 - x \\ \\ \end{gathered}
5=14−x
\begin{gathered}\sf \: \implies \: x = 9 \\ \\ \end{gathered}
⟹x=9
On substituting x = 9, in equation (1), we get
\begin{gathered}\sf \: 9 + y = 24 \\ \\ \end{gathered}
9+y=24
\begin{gathered}\sf \: \implies \: y = 15 \\ \\ \end{gathered}
⟹y=15
\rule{190pt}{2pt}
Additional Information :-
1. Mean using Direct Method
\begin{gathered}\boxed{ \rm{ \:Mean = \dfrac{ \sum f_i x_i}{ \sum f_i} \: }} \\ \\ \end{gathered}
Mean=
∑f
i
2. Mean using Short Cut Method
\begin{gathered}\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i} \: }} \\ \\ \end{gathered}
Mean=A+
d
3. Mean using Step Deviation Method
\begin{gathered}\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h \: }} \\ \\ \end{gathered}
u
×h
Answer:
Hello Refers To An Expression Or Gesture Of Greeting. It's Mainly Used Interjectionally In Greetings, Answering Phone Call Or To Express Surprise.
Eg -- Hello Buddy..!!
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Explanation:
\large\underline{\sf{Solution-}}
Solution−
Calculations of Median
\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \small\boxed{\begin{array}{c |c|c} \tt{Classes} & \tt{Frequency}& \tt{c.f.} \\ \dfrac{\qquad\qquad}{ \sf 0-100} &\dfrac{\qquad\qquad}{ \sf 2} &\dfrac{\qquad\qquad}{ \sf 2}& \\ \dfrac{\qquad\qquad}{ \sf 100-200} &\dfrac{\qquad\qquad}{ \sf 5}&\dfrac{\qquad\qquad}{ \sf 7} & \\ \dfrac{\qquad\qquad}{ \sf 200-300} &\dfrac{\qquad\qquad}{ \sf x}&\dfrac{\qquad\qquad}{ \sf 7 + x} & \\ \dfrac{\qquad\qquad}{ \sf 300-400} &\dfrac{\qquad\qquad}{ \sf 12}&\dfrac{\qquad\qquad}{ \sf 19 + x} & \\ \dfrac{\qquad\qquad}{ \sf 400-500} &\dfrac{\qquad\qquad}{ \sf 17}&\dfrac{\qquad\qquad}{ \sf 36 + x} & \\ \dfrac{\qquad\qquad}{ \sf 500-600} &\dfrac{\qquad\qquad}{ \sf 20}&\dfrac{\qquad\qquad}{ \sf 56 + x} & \\ \dfrac{\qquad\qquad}{ \sf 600-700} &\dfrac{\qquad\qquad}{ \sf y}&\dfrac{\qquad\qquad}{ \sf 56 + x + y}\\ \dfrac{\qquad\qquad}{ \sf 700-800} &\dfrac{\qquad\qquad}{ \sf 9}&\dfrac{\qquad\qquad}{ \sf 65 + x + y}\\ \dfrac{\qquad\qquad}{ \sf 800-900} &\dfrac{\qquad\qquad}{ \sf 7}&\dfrac{\qquad\qquad}{ \sf 72 + x + y}\\ \dfrac{\qquad\qquad}{ \sf 900-1000} &\dfrac{\qquad\qquad}{ \sf 4}&\dfrac{\qquad\qquad}{ \sf 76 + x + y} &\end{array}} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \\ \end{gathered}
Classes
0−100
100−200
200−300
300−400
400−500
500−600
600−700
700−800
800−900
900−1000
Frequency
2
5
x
12
17
20
y
9
7
4
c.f.
2
7
7+x
19+x
36+x
56+x
56+x+y
65+x+y
72+x+y
76+x+y
We have given that,
\begin{gathered}\sf \: \sum \: f \: = \: 100 \\ \\ \end{gathered}
∑f=100
\begin{gathered}\sf \: 76 + x + y= \: 100 \\ \\ \end{gathered}
76+x+y=100
\begin{gathered}\sf \: x + y= \: 100 - 76\\ \\ \end{gathered}
x+y=100−76
\begin{gathered}\sf \: \sf \: \implies \: x + y= \: 24 - - - (1)\\ \\ \\ \end{gathered}
⟹x+y=24−−−(1)
Now, we know
\begin{gathered}\sf \: \sf \: \boxed{ \sf Median= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}} \\ \\ \end{gathered}
Median=l+{h×
f
(
2
N
−cf)
}
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
So, from above calculations, we have
Median = 525
Median class = 500 - 600
l = 500,
h = 100,
f = 20,
cf = cf of preceding class = 36 + x
N = 100
By substituting all the given values in the formula,
\begin{gathered}\sf \: Median= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \} \\ \\ \end{gathered}
Median=l+{h×
f
(
2
N
−cf)
}
\begin{gathered}\sf \: 525= 500 + \Bigg \{100\times \dfrac{ \bigg( 50 - (36 - x)\bigg)}{20} \Bigg \} \\ \\ \end{gathered}
525=500+{100×
20
(50−(36−x))
}
\begin{gathered}\sf \: 525 - 500 = 5\times (50 - 36 - x) \\ \\ \end{gathered}
525−500=5×(50−36−x)
\begin{gathered}\sf \: 25 = 5\times (14 - x) \\ \\ \end{gathered}
25=5×(14−x)
\begin{gathered}\sf \: 5 = 14 - x \\ \\ \end{gathered}
5=14−x
\begin{gathered}\sf \: \implies \: x = 9 \\ \\ \end{gathered}
⟹x=9
On substituting x = 9, in equation (1), we get
\begin{gathered}\sf \: 9 + y = 24 \\ \\ \end{gathered}
9+y=24
\begin{gathered}\sf \: \implies \: y = 15 \\ \\ \end{gathered}
⟹y=15
\rule{190pt}{2pt}
Additional Information :-
1. Mean using Direct Method
\begin{gathered}\boxed{ \rm{ \:Mean = \dfrac{ \sum f_i x_i}{ \sum f_i} \: }} \\ \\ \end{gathered}
Mean=
∑f
i
∑f
i
x
i
2. Mean using Short Cut Method
\begin{gathered}\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i} \: }} \\ \\ \end{gathered}
Mean=A+
∑f
i
∑f
i
d
i
3. Mean using Step Deviation Method
\begin{gathered}\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h \: }} \\ \\ \end{gathered}
Mean=A+
∑f
i
∑f
i
u
i
×h
Verified answer
Answer:
Hello Refers To An Expression Or Gesture Of Greeting. It's Mainly Used Interjectionally In Greetings, Answering Phone Call Or To Express Surprise.
Eg -- Hello Buddy..!!