 # 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

Question 10 Mean Median And Mode Of Ungrouped Data Exercise 18B

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

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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