If the number of scores is odd, then the ( \(\frac{n+1}{2}\)) score is

1.	If the number of scores is odd, then the ( \(\frac{n+1}{2}\)) score is the median of the data. That is, the number of scores below as well as above K\(\frac{n+1}{2}\) is \(\frac{n- 1}{2}\) Verify the fact by taking n = 2m + I.
Answer» proof: Given that, n = 2m + 1 ∴ The sequence of the terms of scores can be 1, 2, 3, ...., 2m + 1 Here, (n +1)/2 = (2m + 1 + 1)/2 = (2m + 2)/2 = 2(m + 1)/2 = m + 1 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 + (n₁ – 1) d ∴ m = 1 + (n₁ – 1)1 ∴ m = 1 + n₁ – 1 ∴ m = n₁ 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 + (n₂ – 1)d ∴ 2m + 1 = m + 2 + (n₂ – 1)1 ∴ 2m + 1 = m + n₂ + 1 ∴ m = n₂ ∴ number of terms from 1 to m = number of terms from m + 2 to 2m + 1 = m = (n-1)/2 ∴ m + 1 is the middle term if the number of scores is 2m + 1.

If the number of scores is odd, then the ( \(\frac{n+1}{2}\)) score is the median of the data. That is, the number of scores below as well as above K\(\frac{n+1}{2}\) is \(\frac{n- 1}{2}\) Verify the fact by taking n = 2m + I.

Answer»

proof:

Given that, n = 2m + 1

∴ The sequence of the terms of scores can be 1, 2, 3, ...., 2m + 1

Here, (n +1)/2 = (2m + 1 + 1)/2

= (2m + 2)/2 = 2(m + 1)/2

= m + 1

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 + (n₁ – 1) d

∴ m = 1 + (n₁ – 1)1

∴ m = 1 + n₁ – 1

∴ m = n₁

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 + (n₂ – 1)d

∴ 2m + 1 = m + 2 + (n₂ – 1)1

∴ 2m + 1 = m + n₂ + 1

∴ m = n₂

∴ number of terms from 1 to m = number of terms from m + 2 to 2m + 1 = m = (n-1)/2

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