# Ex.14.1 Q3 Statistics Solution - NCERT Maths Class 10

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## Question

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs $$18$$. Find the missing frequency $$f$$.

 Daily pocket allowance (in Rs) $$11 – 13$$ $$13 – 15$$ $$15 – 17$$ $$17 – 19$$ $$19 – 21$$ $$21 – 23$$ $$23 – 25$$ Number of children $$7$$ $$6$$ $$9$$ $$13$$ $$f$$ $$5$$ $$4$$

## Text Solution

What is known?

The mean pocket allowance is Rs $$18$$.

What is unknown?

The missing frequency $$f$$.

Reasoning:

We will use assumed mean method to solve this question

\begin{align} \operatorname{Mean,}\,\,\,(\overline{{x}})={a}+\left(\frac{\Sigma {f}_{{i}} {d}_{{i}}}{\Sigma {f}_{{i}}}\right) \end{align}

Solution:

We know that,

$\text{Class mark}, x_i = \frac{{\rm Upper \;class \;limit\; + \;Lower \;class \;limit}}{2}$

Taking assumed mean$$, a = 18$$

Steps:

 Daily pocket allowance (in Rs) No of children $$(f_i)$$ $$X_i$$ $$d_i = x_i -a$$ $$f_id_i$$ $$11 – 13$$ $$7$$ $$12$$ $$-6$$ $$-42$$ $$13 – 15$$ $$6$$ $$14$$ $$-4$$ $$-24$$ $$15 – 17$$ $$9$$ $$16$$ $$-2$$ $$-18$$ $$17 – 19$$ $$13$$ $$18 (a)$$ $$0$$ $$0$$ $$19 – 21$$ $$f$$ $$20$$ $$2$$ $$2f$$ $$21 – 23$$ $$5$$ $$22$$ $$4$$ $$20$$ $$23 – 25$$ $$4$$ $$24$$ $$6$$ $$24$$ $$\Sigma f_i=40 +f$$ $$\Sigma f_id_i=2f-40$$

From the table, we obtain

$$\Sigma {f}_{i} = 40 + f$$

$$\Sigma {f}_{i} {d}_{i} = 2f - 40$$

\begin{align} \text { Mean, }\,\,(\overline{{x}}) &={a}+\left(\frac{\Sigma {f}_{{i}} {d}_{{i}}}{\Sigma {f}_{{i}}}\right) \\ 18 &=18+\left(\frac{-40+2 {f}}{40+f}\right) \\ 18-18 &=\frac{2 {f}-40}{40+f} \\ 2f - 40 &= 0 \\ {f} &=20 \end{align}

Hence, the missing frequency $$f$$ = $$20$$.

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