Sir I have a question to ask from you of statistics..... Find the missing frequencies in the following distribution, if the sum of frequencies is 120 and the mean is 50. Class 0 - 20 frequency 17 class 20 - 40 frequency F1 class 40 - 60 frequency 32 class 60 - 80 frequency F2 class 80 - 100 frequency 19
To solve this, we will use the information provided and the formula for the mean of grouped data Where: 𝑓 f is the frequency of the class. 𝑥 x is the midpoint of the class interval. 𝑁 N is the total sum of frequencies. Step 1: Determine the midpoints of each class: For the class 0 − 20 0−20, midpoint 𝑥 1 = 0 + 20 2 = 10 x 1 = 2 0+20 =10 For the class 20 − 40 20−40, midpoint 𝑥 2 = 20 + 40 2 = 30 x 2 = 2 20+40 =30 For the class 40 − 60 40−60, midpoint 𝑥 3 = 40 + 60 2 = 50 x 3 = 2 40+60 =50 For the class 60 − 80 60−80, midpoint 𝑥 4 = 60 + 80 2 = 70 x 4 = 2 60+80 =70 For the class 80 − 100 80−100, midpoint 𝑥 5 = 80 + 100 2 = 90 x 5 = 2 80+100 =90 Step 2: Set up the sum of frequencies and the sum of 𝑓 ⋅ 𝑥 f⋅x: We know that: The total sum of frequencies is 120. 17 + 𝐹 1 + 32 + 𝐹 2 + 19 = 120 17+F 1 +32+F 2 +19=120 Simplifying: 𝐹 1 + 𝐹 2 = 52 F 1 +F 2 =52 The total sum of 𝑓 ⋅ 𝑥 f⋅x is the mean times the total frequencies: 50 × 120 = 6000 50×120=6000 Now calculate the sum of 𝑓 ⋅ 𝑥 f⋅x for the known frequencies: 17 × 10 = 170 17×10=170 32 × 50 = 1600 32×50=1600 19 × 90 = 1710 19×90=1710 Let’s write the equation for the sum of 𝑓 ⋅ 𝑥 f⋅x: 170 + 𝐹 1 × 30 + 1600 + 𝐹 2 × 70 + 1710 = 6000 170+F 1 ×30+1600+F 2 ×70+1710=6000 Simplifying: 𝐹 1 × 30 + 𝐹 2 × 70 + 3480 = 6000 F 1 ×30+F 2 ×70+3480=6000 𝐹 1 × 30 + 𝐹 2 × 70 = 2520 F 1 ×30+F 2 ×70=2520 Step 3: Solve the system of equations: We now have the system of two equations: 𝐹 1 + 𝐹 2 = 52 F 1 +F 2 =52 30 𝐹 1 + 70 𝐹 2 = 2520 30F 1 +70F 2 =2520 We can solve this system by substitution or elimination. Let's multiply the first equation by 30: 30 𝐹 1 + 30 𝐹 2 = 1560 30F 1 +30F 2 =1560 Now subtract this from the second equation: ( 30 𝐹 1 + 70 𝐹 2 ) − ( 30 𝐹 1 + 30 𝐹 2 ) = 2520 − 1560 (30F 1 +70F 2 )−(30F 1 +30F 2 )=2520−1560 40 𝐹 2 = 960 40F 2 =960 𝐹 2 = 24 F 2 =24 Now substitute 𝐹 2 = 24 F 2 =24 into the first equation: 𝐹 1 + 24 = 52 F 1 +24=52 𝐹 1 = 28 F 1 =28 Final Answer: 𝐹 1 = 28 F 1 =28 𝐹 2 = 24 F 2 =24 Thus, the missing frequencies are 𝐹 1 = 28 F 1 =28 and 𝐹 2 = 24 F 2 =24.
wish all the ninethies a very best for your final exams. May we all score our desired marks in our exams best of luck for your upcoming exams we all will make it
Just an extra point : In rs Agarwal it is written that when the degree of a polynomial is 4 then the polynomial is called "biquadratic" It was not mentioned in the video so I thought I should tell you guys
Sir #8 question me ham agar dono equations ko or solve kare to Equation 1 =4p+r=-10 Or Equation 2 =p+4r=-10 hoga To dono ke rhs equal he So on equalising the LHS WE get 4p+r=p+4r =4p-p=4r-r So,3p=3r And 3 wil get cancelled out from both So,p=r An easier way to do the same question😊
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Sir I have a question to ask from you of statistics.....
Find the missing frequencies in the following distribution, if the sum of frequencies is 120 and the mean is 50. Class 0 - 20 frequency 17 class 20 - 40 frequency F1 class 40 - 60 frequency 32 class 60 - 80 frequency F2 class 80 - 100 frequency 19
To solve this, we will use the information provided and the formula for the mean of grouped data
Where:
𝑓
f is the frequency of the class.
𝑥
x is the midpoint of the class interval.
𝑁
N is the total sum of frequencies.
Step 1: Determine the midpoints of each class:
For the class
0
−
20
0−20, midpoint
𝑥
1
=
0
+
20
2
=
10
x
1
=
2
0+20
=10
For the class
20
−
40
20−40, midpoint
𝑥
2
=
20
+
40
2
=
30
x
2
=
2
20+40
=30
For the class
40
−
60
40−60, midpoint
𝑥
3
=
40
+
60
2
=
50
x
3
=
2
40+60
=50
For the class
60
−
80
60−80, midpoint
𝑥
4
=
60
+
80
2
=
70
x
4
=
2
60+80
=70
For the class
80
−
100
80−100, midpoint
𝑥
5
=
80
+
100
2
=
90
x
5
=
2
80+100
=90
Step 2: Set up the sum of frequencies and the sum of
𝑓
⋅
𝑥
f⋅x:
We know that:
The total sum of frequencies is 120.
17
+
𝐹
1
+
32
+
𝐹
2
+
19
=
120
17+F
1
+32+F
2
+19=120
Simplifying:
𝐹
1
+
𝐹
2
=
52
F
1
+F
2
=52
The total sum of
𝑓
⋅
𝑥
f⋅x is the mean times the total frequencies:
50
×
120
=
6000
50×120=6000
Now calculate the sum of
𝑓
⋅
𝑥
f⋅x for the known frequencies:
17
×
10
=
170
17×10=170
32
×
50
=
1600
32×50=1600
19
×
90
=
1710
19×90=1710
Let’s write the equation for the sum of
𝑓
⋅
𝑥
f⋅x:
170
+
𝐹
1
×
30
+
1600
+
𝐹
2
×
70
+
1710
=
6000
170+F
1
×30+1600+F
2
×70+1710=6000
Simplifying:
𝐹
1
×
30
+
𝐹
2
×
70
+
3480
=
6000
F
1
×30+F
2
×70+3480=6000
𝐹
1
×
30
+
𝐹
2
×
70
=
2520
F
1
×30+F
2
×70=2520
Step 3: Solve the system of equations:
We now have the system of two equations:
𝐹
1
+
𝐹
2
=
52
F
1
+F
2
=52
30
𝐹
1
+
70
𝐹
2
=
2520
30F
1
+70F
2
=2520
We can solve this system by substitution or elimination. Let's multiply the first equation by 30:
30
𝐹
1
+
30
𝐹
2
=
1560
30F
1
+30F
2
=1560
Now subtract this from the second equation:
(
30
𝐹
1
+
70
𝐹
2
)
−
(
30
𝐹
1
+
30
𝐹
2
)
=
2520
−
1560
(30F
1
+70F
2
)−(30F
1
+30F
2
)=2520−1560
40
𝐹
2
=
960
40F
2
=960
𝐹
2
=
24
F
2
=24
Now substitute
𝐹
2
=
24
F
2
=24 into the first equation:
𝐹
1
+
24
=
52
F
1
+24=52
𝐹
1
=
28
F
1
=28
Final Answer:
𝐹
1
=
28
F
1
=28
𝐹
2
=
24
F
2
=24
Thus, the missing frequencies are
𝐹
1
=
28
F
1
=28 and
𝐹
2
=
24
F
2
=24.
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wish all the ninethies a very best for your final exams. May we all score our desired marks in our exams
best of luck for your upcoming exams
we all will make it
1:00:50 sir ki editing
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Just an extra point :
In rs Agarwal it is written that when the degree of a polynomial is 4 then the polynomial is called "biquadratic"
It was not mentioned in the video so I thought I should tell you guys
Thanks ❤
L
thank you!!!!!!!
Take my L
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Aaj polynomials ka revision ho jayega, thank you bhaiyaaa ❤🥰
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Sir
#8 question me ham agar dono equations ko or solve kare to
Equation 1 =4p+r=-10
Or
Equation 2 =p+4r=-10 hoga
To dono ke rhs equal he
So on equalising the LHS
WE get
4p+r=p+4r
=4p-p=4r-r
So,3p=3r
And 3 wil get cancelled out from both
So,p=r
An easier way to do the same question😊
Yes
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