**Chapter 6 Statistics Set 6.2**

**Chapter 6 Statistics Set 6.2**

**Question 1.The following table shows classification of number of workers and the number of hours they work in a software company. Find the median of the number of hours they work.**

Solution:

Solution:

Cumulative frequency which is just greater than (or equal) to 500 is 650.

∴ The median class is 10 – 12.

Now, L = 10, f = 500, cf = 150, h = 2

∴ The median of the number of hours the workers work is 11.4 hours.

**Question 2.The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data.Solution:**

Here, total frequency = ∑f

_{i}= N = 250

Cumulative frequency which is just greater than (or equal) to 125 is 153.

∴ The median class is 150 – 200.

Now, L = 150, f = 90, cf = 63, h = 50

∴ The median of the given data is 184 mangoes (approx).

**Question 3.The following table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data.Solution:**

Here, total frequency = ∑f

_{i}= N = 200

Cumulative frequency which is just greater than (or equal) to 100 is 184.

∴ The median class is 74.5 – 79.5.

Now, L = 74.5, f = 85, cf = 99, h = 5

∴ The median of the given data is 75 km/hr (approx.).

**Question 4.The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.Solution:**

Cumulative frequency which is just greater than (or equal) to 52.5 is 67.

∴ The median class is 50 – 60.

Now, L = 50, f = 20, cf = 47, h = 10

∴ The median of the productions is 52750 bulbs (approx.).

**Question 1.Solution:**

The sequence of the terms of scores is 1,2, 3, …….., m, m + 1, m + 2, …, 2m + 1

Thus, we have to prove that m + 1 is the middle term if the number of scores is 2m + 1

i.e. to prove

number of terms from 1 to m = number of terms from m + 2 to 2m + 1 …(i)

Consider the L.H.S. of equation (i)

The sequence is an A.P. with a = 1,d = 1, t

_{n1}= m

t

_{n1}= a + (n

_{1}– 1) d

∴ m = 1 + (n

_{1}– 1)1

∴ m = 1 + n

_{1}– 1

∴ m = n

_{1}

Consider the R.H.S. of equation (ii)

The sequence is an A.P. with a = m + 2, d = 1, t

_{n2}= 2m + 1

t

_{n2}= a + (n

_{2}– 1)d

∴ 2m + 1 = m + 2 + (n

_{2}– 1)1

∴ 2m + 1 = m + n

_{2}+ 1

∴ m = n

_{2}

∴ number of terms from 1 to m = number of terms from m + 2 to 2m + 1 = m =

∴ m + 1 is the middle term if the number of scores is 2m + 1.

Solution:

The sequence of the terms of scores is 1, 2, 3 … m – 1, m, m + 1, m + 2,…., 2m

Thus, we have to prove that m and m + 1 are the middlemost terms if the number of scores is 2m.

i.e. to prove

number of terms from 1 to m – 1 = number of terms from m + 2 to 2m …(i)

Consider the L.H.S. of equation (i)

The sequence is an A.P. with a = 1, d = 1, t_{n1} = m – 1

t_{n1} = a + (n_{1} – 1)d

∴ m – 1 = 1 + (n_{1} – 1)1

∴m – 1 = 1 + n_{1} – 1

∴ n_{1} = m – 1

Consider the R.H.S. of equation (i)

The sequence is an A.P. with a = m + 2, d = 1, t_{n2}= 2m

t_{n2}= a + (n_{2} – 1) d

∴ 2m = m + 2 + (n_{2} – 1)1

∴ 2m = m + 2 + n_{2} – 1

∴ n_{2} = m – 1

∴ number of terms from 1 to m – 1 = number of terms from m + 2 to 2m = m – 1 =

∴ m and m + 1 are the middlemost terms if the number of scores is 2m.