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RS Aggarwal Solutions Class 9 Mathematics Solutions for Mean Median And Mode Of Ungrouped Data Exercise 18B in Chapter 18 - Mean Median And Mode Of Ungrouped Data
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Find the value of p for the following frequency distribution whose mean is 16.6.
| x | f |
|---|---|
| 8 | 12 |
| 12 | 16 |
| 15 | 20 |
| p | 24 |
| 20 | 16 |
| 25 | 8 |
| 30 | 4 |
Answer:
| x | f | xifi |
|---|---|---|
| 8 | 12 | 96 |
| 12 | 16 | 192 |
| 15 | 20 | 300 |
| p | 24 | 24p |
| 20 | 16 | 320 |
| 25 | 8 | 200 |
| 30 | 4 | 120 |
| Σ fi = 100 | Σ fi xi = 1228 + 24p |
It is given that
Mean = 16.6
We know that
Mean = Σ fi xi/ Σ fi
So we get
(1228 + 24p)/ 100 = 16.6
On further calculation
1228 + 24p = 1660
So we get
24p = 1660 – 1228
By subtraction
24p = 432
We get
p = 18
Therefore, the value of p is 18.
Video transcript
"hello students welcome back to leader
homework
my name is anand anand sharma question
is
find the value of p for the following
frequency distribution whose mean is
16.6 so according to given information
we have to prepare a table
note down the x values given in the
question
these are 8 12
15 p
20 25
and 30. frequencies respective
frequencies are
12 16
20 24
16 8 and 4
we need to find the product of x and f
and we will denote it x i f i
so first product is
96 192
300 24
p 320
200 and 120 last now
we will add up all the frequencies that
is column number two
and we'll denote it summation fi which
is
100 and we will add x i f i column as
well
and that will denote summation
f i x i and this all addition will be
1228 plus 24p
it is given that frequency distribution
mean
is 16.6
so you can note down here
mean is 16.6 according to the formula of
the mean
we have mean is equal to summation f
i x i upon summation f i
this is the formula so we substitute the
values here in this
we do get 1 2
2 8 plus 24 p
upon 100 equals
16.6 cross multiplication that is
100 shifting on that side you will get
1 2 2 8 plus
24 p is equal to 1
6 6 0 and further calculation will give
you
p value as 18
so this unknown p value is 18.
that's all for this question if you have
any queries feel free to comment below
and don't forget to subscribe to the
channel for
all the notifications of further videos
thank you
"
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