Statistics

NCERT Exercise 14.3

Question 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.

Answer:

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:

`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.

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

Answer: `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

Question 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.

Answer:

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`

Question 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:

Find the median length of leaves.

Answer:

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`

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

Find the median life time of a lamp.

Answer:

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`

Question 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:

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.

Answer: Calculations for median:

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:

`x= (Σf_i\x_i)/(Σf_i)`

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

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

Answer:

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`