[Solved] What Is The Value Of Q ?

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Concept:

If x1, x2, x3, …xn are the number of observations (Or class marks of classes) with respective frequencies f1, f2, f3, … fn, then

The sum of observations = f1x1 + f2x2 + f3x3 +….+ fnxn

The total number of observations = f1+ f2 +…..+ fn

Therefore, the mean of the data, x̄ = (f1x1+ f2x2 + f3x3 + ….+ fnxn)/( f1+f2+… + fn).

The mean of the data, x̅ = \(\displaystyle ∑\frac{f_ix_i}{N} \)

Where, xi is the class marks of classes

Class mark = (lower limit + upper limit)/2

Calculation:

Total frequency = 120

⇒ 17 + p + q + 32 + p - 3q + 19 = 120

⇒ 2p - 2q = 52

⇒ p - q = 26 ------(1)

We know mean x̅ = \(\displaystyle ∑\frac{f_ix_i}{N} \),

⇒ Class mark of 1st class intreval (x1)= \(\displaystyle \frac{x_l + x_u}{2}\) = \(\displaystyle \frac{0 + 20}{2}\) = 10

Similarly, we get

⇒ x2 = \(\displaystyle \frac{20+40}{2}\) = 30

⇒ x3 = \(\displaystyle \frac{40 + 60}{2}\) = 50

⇒ x4 = \(\displaystyle \frac{60 + 80}{2}\) = 70

⇒ x5 = \(\displaystyle \frac{80 + 100}{2}\) = 90

⇒ x̅ = \(\displaystyle ∑\frac{f_ix_i}{N} \)

⇒ x̅ = \(\displaystyle \frac{f_1x_1+f_2x_2+f_3x_3+f_4x_4+f_5x_5}{N}\)

⇒ 50 = \(\displaystyle \frac{17×10+(p+q)×30+32×50+(p-3q)×70+19×90}{120}\)

⇒ 6000 = 170 +30p + 30q + 1600 + 70p - 210 q + 1710

⇒ 6000 = 100p - 180q + 3480

⇒ 2520 = 100p - 180q

⇒ 252 = 10p - 18q ------(2)

Solving (1) and (2) we get

⇒ q = 1

∴ The value of q = 1.

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