Class 10 Mathematics

Statistics

Exercise 14.3(NCERT)

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.

Monthly consumption (in units)Number of customers
65-854
85-1055
105-12513
125-14520
145-16514
165-1858
185-2054

Solution:

Class IntervalFrequencyCumulative frequency
65-8544
85-10559
105-1251322
125-1452042
145-1651456
165-185864
185-205468
N = 68

Here; n = 68 and hence `n/2 = 34`

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

now, `l = 125`, `n = 68`, `cf = 22`, `f = 20`, `h = 20`

Median can be calculated as follows:

`text(Median)=l+(n/2-cf)/(f)xxh`

`=125+(34-22)/(20)xx20`

`=125+12=137`

Calculations for Mode:

Modal class `= 125-145`, `f_1 = 20`, `f_0 = 13`, `f_2 = 14` and `h = 20`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=125+(20-13)/(2xx20-13-14)xx20`

`=125+7/13xx20`

`=125+10.77=135.77`

Calculations for Mean:
Class Intervalfixidi = xi - aui = di/hfiui
65-85475-60-3-12
85-105595-40-2-10
105-12513115-20-1-13
125-14520135000
145-1651415520114
165-185817540216
185-205419560312
Σ fi = 68Σ fiui = 7

`x=a+(Σf_i\u_i)/(Σf_i)xxh`

`=135+7/68xx20=137.05`

Mean, median and mode are more or less equal in this distribution.


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

Class IntervalFrequency
0-105
10-20x
20-3020
30-4015
40-50y
50-605
Total60

Solution: `n = 60` and hence `n/2 = 30`

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

lower limit of median class = 20, `cf = 5 + x`, `f = 20` and `h = 10`

`text(Median)=l+(n/2-cf)/(f)xxh`

Or, `28.5=20+(30-5-x)/(20)xx10`

Or, `(25-x)/(2)=8.5`

Or, `25-x=17`

Or, `x=25-17=18`

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

`60=5+20+15+5+x+y`
Or, `45+x+y=60`
Or, `x+y=60-45=15`
Hence, `y=15-x=15-8=7`

Hence, x = 8 and y = 7

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

Age (in years)Number of policyt hodlers
Below 202
Below 256
Below 3024
Below 3545
Below 4078
Below 4589
Below 5092
Below 5598
Below 60100

Solution:

Class intervalFrequencyCumulative frequency
15-2022
20-2546
25-301824
30-352145
35-403378
40-451189
45-50392
50-55698
55-602100

Here; `n = 100` and `n/2 = 50`, hence median class `= 35-45`

In this case; `l = 35`, `cf = 45`, `f = 33` and `h = 5`

`text(Median)=l+(n/2-cf)/(f)xxh`

`=35+(50-45)/(33)xx5`

`=35+25/33=35.75`

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

Length (in mm)Number of leaves
118-1263
127-1355
136-1449
145-15312
154-1625
163-1714
172-1802

Find the median length of leaves.


Solution:

Class IntervalFrequencyCumulative frequency
117.5-126.533
126.5-135.558
135.5-144.5917
144.5-153.51229
153.5-162.5534
162.5-171.5438
171.5-180.5240

We have; `n = 40` and `n/2 = 20` hence median class `= 144.5-153.5`

Thus, `l = 144.5`, `cf = 17`, `f = 12` and `h = 9`

`text(Median)=l+(n/2-cf)/(f)xxh`

`=144.5+(20-17)/(12)xx9`

`=144.5+9/4=146.75`

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

Lifetime (in hours)Number of lamps
1500-200014
2000-250056
2500-300060
3000-350086
3500-400074
4000-450062
4500-500048

Find the median life time of a lamp.

Solution:

Class IntervalFrequencyCumulative Frequency
1500-20001414
2000-25005670
2500-300060130
3000-350086216
3500-400074290
4000-450062352
4500-500048400

We have; `n = 400` and `n/2 = 200` hence median class `= 3000 – 3500`

So, `l = 3000`, `cf = 130`, `f = 86` and `h = 500`

`text(Median)=l+(n/2-cf)/(f)xxh`

`=3000+(200-130)/(86)xx500`

`=3000+406.97=3406.97`

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

Number of letters1-44-77-1010-1313-1616-19
Number of surnames630401644

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: Calculations for median:

Class IntervalFrequencyCumulative Frequency
1-466
4-73036
7-104076
10-131692
13-16496
16-194100

Here; `n = 100` and `n/2 = 50` hence median class `= 7-10`

So, `l = 7`, `cf = 36`, `f = 40` and `h = 3`

`text(Median)=l+(n/2-cf)/(f)xxh`

`=7+(50-36)/(40)xx3`

`=7+14/40xx3=8.05`

Calculations for Mode:

Modal class `= 7-10`

Here; `l = 7`, `f_1 = 40`, `f_0 = 30`, `f_2 = 16` and `h = 3`

Mode `=l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

`=7+(40-30)/(2xx40-30-16)xx3`

`=7+10/34xx3=7.88`

Calculations for Mean:
Class intervalfixifixi
1-462.515
4-7305.5165
7-10408.5340
10-131611.5184
13-16414.551
16-19417.570
Σ fi = 100Σ fixi = 825

`x= (Σf_i\x_i)/(Σf_i)`

`=(825)/(100)=8.25`

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

Weight (in kg)40-4545-5050-5555-6060-6565-7070-75
Number of students2386632

Solution:

Class IntervalFrequencyCumulative frequency
40-4522
45-5035
50-55813
55-60619
60-65625
65-70328
70-75230

We have; `n = 30` and `n/2 = 15` hence median class `= 55-60`

So, `l = 55`, `cf = 13`, `f = 6` and `h = 5`

`text(Median)=l+(n/2-cf)/(f)xxh`

`=55+(15-13)/(6)xx5`

`=55+5/3=56.67`


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Exercise 1

Exercise 2