Exercise-3

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1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean, and mode of the data and compare them.

Solution:

Mean of the data:

assumed mean (a)=135

class size (h) = 205-185=20

Mean(`\bar x`) = `a+ {sum f_{i}u_{i}}/{sum f_{i}}times h `

`\therefore \bar x = 135+ {7}/{68}times 20 `

= 135+`140/68`=135+2.058=137.058.

`therefore` mean of the data is 137.1

Median of the data:

Total n=68 `\Rightarrow n/2=68/2=34`

cumulative frequency just greater than 34 is 42 it is belonging to the interval 125-145.

So, the median class is 125-145

Lower boundary `l` of median class=125

Class size (`h`) = 20

Frequency (`f`) of median class = 20

Cumulative frequency (`cf`) of class preceding median class = 22

Median = `l+[(n/2-cf)/f]times h`

  = `125+[(34-22)/20]times 20= 125 + 12=137`

∴ median of the data is 137

Mode of the data:

Since the maximum number of consumers (i.e., 20) is in the class interval 125-145

So, the modal class is 125-145

The lower boundary (` l `) of the modal class =125,

The class size ( `h`) = 20,

The frequency of modal class ( `f_1 `) = 20,

the frequency of the class preceding the modal class ( `f_0` ) = 13,

the frequency of the class succeeding the modal class ( `f_2` ) = 14.

Mode `=l+[{f_1-f_0}/{2f_1-f_0-f_2}]times h`

`=125+[{20-13}/{2times20-13-14}]times 20`

=125+`140/13 `=125+10.769=135.769

`\therefore`Mode of given data is 135.8.

For given data Mean=137.1,Median=137 and Mode =135.8 are approximately equal.

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2. If the median of 60 observations, given below is 28.5, find the values of x and y.

Solution:

Median =28.5

∴ 20 - 30 is median class

Then, `l` (the lower boundary) = 20,

cf (the cumulative frequency of the class preceding 20-30) = 5+x,

f (the frequency of the median class 20-30) = 20,

`h` (the class size) = 10

Median = `l+[(n/2-cf)/f]times h`

28.5`= 20+[(30-(5+x))/20]times 10`

`28.5-20= [(30-5-x)/2]`

`8.5 = [(25-x)/2]`

`8.5 times 2 = 25-x`

∴ x = 25-17 = 8

Given sum of frequency = 60

∴ 45+x+y=60

45+8+y=60

y`= 60-53=7`

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3. A life insurance agent found the following data about the distribution of ages of 100 policyholders. Calculate the median age.

[Policies are given only to persons having age 18 years onwards but less than 60 years.]

Solution:

As policies were given only to persons having age 18 years onwards but less than 60 years we can definitely class intervals with their respective cumulative Frequency as below

Total n=100 `\Rightarrow n/2=100/2=50`

cumulative frequency just greater than 50 is 78 it is belonging to the interval 35-40.

So, the median class is 35-40

Lower boundary `l` of median class=35

Class size (`h`) = 5

Frequency (`f`) of median class = 33

Cumulative frequency (`cf`) of class preceding median class = 45

Median = `l+[(n/2-cf)/f]times h`

Median = `35+[(100/2-45)/33]times 5`

= 35 + `(50-45)/33 times 5`

= 35 + `25/33`=35+0.7575=35.76

∴The median age of policyholders is 𝟑𝟓.𝟕𝟔 years

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4. The lengths of 40 leaves of a plant are measured correctly to the nearest millimeter, and the data obtained is represented in the following table Find the median length of the leaves.

Solution:

The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.

n= 40 `Rightarrow n/2=40/2=20`

From the table, we may observe that cumulative Frequency just greater than 20 is 29

Median class 144.5-153.5

Lower limit (`l`) of median class 144.5

Class size (`h`) =9

Frequency (`f`) of median class = 12

Cumulative frequency (`cf`) of class preceding median class = 17

Median = `l+[(n/2-cf)/f]times h`

Median = `144.5+[(20-17)/12]times 9`

= 144.5 + `9/4`=144.5+2.25=146.75

∴ the median length of leaves is 146.75 mm

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5. The following table gives the distribution of the lifetime of 400 neon lamps. Find the median lifetime of a lamp.

Solution:

n= 400 `Rightarrow n/2=400/2=200`

From the table, we may observe that cumulative Frequency just greater than 200 is 216

Median class 3000-3500

Lower limit (`l`) of median class 3000

Class size (`h`) =500

Frequency (`f`) of median class = 86

Cumulative frequency (`cf`) of class preceding median class = 130

Median = `l+[(n/2-cf)/f]times h`

Median = `3000+[(200-130)/86]times 500`

= 3000 + `35000/86`=3000+406.98=3406.98

∴ median lifetime of a lamp is 3407 hours

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6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows. Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Solution:

Mean of the data:

assumed mean (a)=8.5

class size (h) = 4-1=3

Mean(`\bar x`) = `a+ {sum f_{i}u_{i}}/{sum f_{i}}times h `

`\therefore \bar x = 8.5+ {-6}/{100}times 3 `

= 8.5-0.18=8.32.

`therefore` mean the number of letters in the surnames is 8.32.

Median of the data:

Total n=100 `\Rightarrow n/2=100/2=50`

cumulative frequency just greater than 50 is 76 it is belonging to the interval 7-10.

So, the median class is 7-10

Lower boundary `l` of median class=7

Class size (`h`) = 3

Frequency (`f`) of median class = 40

Cumulative frequency (`cf`) of class preceding median class = 36

Median = `l+[(n/2-cf)/f]times h`

= `7+[(50-36)/40]times 3= 7 + `42/40`=7+1.05=8.05

∴ median number of the letters in the surname is 8.05

Mode of the data:

Since the maximum number of surnames (i.e., 40) is in the class interval 7-10

So, the modal class is 7-10

The lower boundary (` l `) of the modal class =7,

The class size ( `h`) = 3,

The frequency of modal class ( `f_1 `) = 40,

the frequency of the class preceding the modal class ( `f_0` ) = 30,

the frequency of the class succeeding the modal class ( `f_2` ) = 16.

Mode `=l+[{f_1-f_0}/{2f_1-f_0-f_2}]times h`

`=7+[{40-30}/{2times40-30-16}]times 3`

=7+`30/34 `=7+0.88=7.88

`\therefore`Modal size of surnames is 7.88

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7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Solution:

n= 30 `Rightarrow n/2=30/2=15`

From the table, we may observe that cumulative Frequency just greater than 15 is 19

Median class 55-60

Lower limit (`l`) of median class 55

Class size (`h`) =5

Frequency (`f`) of median class = 6

Cumulative frequency (`cf`) of class preceding median class = 13

Median = `l+[(n/2-cf)/f]times h`

Median = `55+[(15-13)/6]times 5`

= 55 + `5/3`=55+1.67=56.67

∴ The median weight of the students is 56.67 kg

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< Median Ogive curves >

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