Question

f the mean of the following distributions 54, find the value of p

Class 60-69 70-79 80-89 90-99 100-109 110-119
Frequency 7 p 1 0 9 13

Answer 1

Answer:

Here, the frequency table is given in inclusive form. So we first transform it into exclusive form by subtracting and adding h/2 to the lower and upper limits respectively of each class, whereh denotes the difference of lower and upper limit of a class and the upper limit of the previous class.

Transforming the above table into exclusive form and preparing the cumulative frequency table, we get

Weekly wages (in Rs.) No. of workers Cumulative frequency

59.5- 69.5 5 5

69.5- 79.5 15 20

79.5- 89.5 20 40

89.5- 99.5 30 70

99.5- 109.5 20 90

109.5- 119.5 8 98

N=∑f

i

=98

We have,

N=98.∴N/2=49

The cumulative frequency just greater than N/2 is 70 and corrssponding class is 89.5−99.5,So,89.5−99.5 is the median class.

∴l=89.5,h=10,f=30,andF=40

Now, Median =l+

f

2

N

−f

×h

⇒Median=89.5+

30

49−40

×10=92.5

Answer 2

Answer:

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4

−14x

2

+7x

\sf{=\dfrac{7x\left(3x^3-2x+1\right)}{7x^3}}=

7x

3

7x(3x

3

−2x+1)

\bf{=\dfrac{3x^3-2x+1}{x^2}}=

x

2

3x

3

−2x+1