Exercise 14.3 Page: 287

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

Solution:

Find the cumulative frequency of the given data as follows:

From the table, it is observed that, n = 68 and hence n/2=34

Hence, the median class is 125-145 with cumulative frequency = 42

Where, l = 125, n = 68, C_f = 22, f = 20, h = 20

Median is calculated as follows:

=125+((34−22)/20) × 20

=125+12 = 137

Therefore, median = 137

To calculate the mode:

Modal class = 125-145,

f₁=20, f₀=13, f₂=14 & h = 20

Mode formula:

Mode = l+ [(f₁-f₀)/(2f₁-f₀-f₂)]×h

Mode = 125 + ((20-13)/(40-13-14))×20

=125+(140/13)

=125+10.77

=135.77

Therefore, mode = 135.77

Calculate the Mean:

x̄ =a+h ∑f_iu_i/∑f_i =135+20(7/68)

Mean=137.05

In this case, mean, median and mode are more/less equal in this distribution.

2. If the median of a distribution given below is 28.5 then, find the value of x & y.

Solution:

Given data, n = 60

Median of the given data = 28.5

Where, n/2 = 30

Median class is 20 – 30 with a cumulative frequency = 25+x

Lower limit of median class, l = 20,

C_f = 5+x,

f = 20 & h = 10

Substitute the values

28.5=20+((30−5−x)/20) × 10

8.5 = (25 – x)/2

17 = 25-x

Therefore, x =8

Now, from cumulative frequency, we can identify the value of x + y as follows:

Since,

60=5+20+15+5+x+y

Now, substitute the value of x, to find y

60 = 5+20+15+5+8+y

y = 60-53

y = 7

Therefore, the value of x = 8 and y = 7.

3. The Life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to the persons whose age is 18 years onwards but less than the 60 years.

Solution:

Given data: n = 100 and n/2 = 50

Median class = 35-45

Then, l = 35, c_f = 45, f = 33 & h = 5

Median = 35+((50-45)/33) × 5

= 35 + (5/33)5

= 35.75

Therefore, the median age = 35.75 years.

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

Find the median length of leaves.             

Solution:

Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

So, the data obtained are:

n = 40 and n/2 = 20

Median class = 144.5-153.5

then, l = 144.5,

cf = 17, f = 12 & h = 9

Median = 144.5+((20-17)/12)×9

= 144.5+(9/4)

= 146.75 mm

Therefore, the median length of the leaves = 146.75 mm.

5. The following table gives the distribution of a life time of 400 neon lamps.

Find the median lifetime of a lamp.

Solution:

Data:

n = 400 &n/2 = 200

Median class = 3000 – 3500

Therefore, l = 3000, C_f = 130,

f = 86 & h = 500

Median = 3000 + ((200-130)/86) × 500

= 3000 + (35000/86)

= 3000 + 406.97

= 3406.97

Therefore, the median life time of the lamps = 3406.97 hours

6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Solution:

To calculate median:

Given:

n = 100 &n/2 = 50

Median class = 7-10

Therefore, l = 7, C_f = 36, f = 40 & h = 3

Median = 7+((50-36)/40) × 3

Median = 7+42/40

Median=8.05

Calculate the Mode:

Modal class = 7-10,

Where, l = 7, f₁ = 40, f₀ = 30, f₂ = 16 & h = 3

Mode = 7+((40-30)/(2×40-30-16)) × 3

= 7+(30/34)

= 7.88

Therefore mode = 7.88

Calculate the Mean:

Mean = x̄ = ∑f_i x_i /∑f_i 

Mean = 825/100 = 8.25

Therefore, mean = 8.25

7. The distributions of below give a weight of 30 students of a class. Find the median weight of a student.

Solution:

Given: n = 30 and n/2= 15

Median class = 55-60

l = 55, C_f = 13, f = 6 & h = 5

Median = 55+((15-13)/6)×5

Median=55 + (10/6) = 55+1.666

Median =56.67

Therefore, the median weight of the students = 56.67