A number of particular articles have been classified according to their weights. After drying for two weeks, the same articles have again been weighed and similarly classified. It is known that the median weight in the first weighing was 20 gm while in the second weighing it was 17 gm. Some frequencies a and b in the first weighing and z and y in the second are missing. It is known that a = 3z and b = 2y.
Answers
Answer:
M_1=20.83 \\ M_2=17.35 \\ x=3a \\ y=2b \\ N_1=a+b+11+52+75+22=160+a+b \\ N_2=x+y+40+50+30+28 = 148 +x+y = 148 +3a+2bM
1
=20.83
M
2
=17.35
x=3a
y=2b
N
1
=a+b+11+52+75+22=160+a+b
N
2
=x+y+40+50+30+28=148+x+y=148+3a+2b
The median class are 20-25 and 10-15 respectively.
Median = l + \frac{n}{f}(\frac{N}{2}+c) \\ M_1= l+ \frac{n}{f}(\frac{N_1}{2} -cf) \\ 20.83 = 20 + \frac{5}{75}[\frac{160+a+b}{2}-(63+a+b)] \\ 12.45 = 17 - \frac{a+b}{2} \\ a+b=9.1 \;(1)\\ M_2 = l + \frac{n}{f}(\frac{N_2}{2} -cf) \\ = 15 + \frac{5}{50}[\frac{148+3a+2b}{2}-(40+3a+2b)] \\ \frac{3a+2b}{2}=10.5 \\ 3a+2b=21 \;(2)Median=l+
f
n
(
2
N
+c)
M
1
=l+
f
n
(
2
N
1
−cf)
20.83=20+
75
5
[
2
160+a+b
−(63+a+b)]
12.45=17−
2
a+b
a+b=9.1(1)
M
2
=l+
f
n
(
2
N
2
−cf)
=15+
50
5
[
2
148+3a+2b
−(40+3a+2b)]
2
3a+2b
=10.5
3a+2b=21(2)
Solving (1) and (2), we get
a=3
b=6
x=9
y=12
Explanation:
M_1=20.83 \\ M_2=17.35 \\ x=3a \\ y=2b \\ N_1=a+b+11+52+75+22=160+a+b \\ N_2=x+y+40+50+30+28 = 148 +x+y = 148 +3a+2bM
1
=20.83
M
2
=17.35
x=3a
y=2b
N
1
=a+b+11+52+75+22=160+a+b
N
2
=x+y+40+50+30+28=148+x+y=148+3a+2b
The median class are 20-25 and 10-15 respectively.
Median = l + \frac{n}{f}(\frac{N}{2}+c) \\ M_1= l+ \frac{n}{f}(\frac{N_1}{2} -cf) \\ 20.83 = 20 + \frac{5}{75}[\frac{160+a+b}{2}-(63+a+b)] \\ 12.45 = 17 - \frac{a+b}{2} \\ a+b=9.1 \;(1)\\ M_2 = l + \frac{n}{f}(\frac{N_2}{2} -cf) \\ = 15 + \frac{5}{50}[\frac{148+3a+2b}{2}-(40+3a+2b)] \\ \frac{3a+2b}{2}=10.5 \\ 3a+2b=21 \;(2)Median=l+
f
n
(
2
N
+c)
M
1
=l+
f
n
(
2
N
1
−cf)
20.83=20+
75
5
[
2
160+a+b
−(63+a+b)]
12.45=17−
2
a+b
a+b=9.1(1)
M
2
=l+
f
n
(
2
N
2
−cf)
=15+
50
5
[
2
148+3a+2b
−(40+3a+2b)]
2
3a+2b
=10.5
3a+2b=21(2)
Solving (1) and (2), we get
a=3
b=6
x=9
y=12