The following distribution shows the daily po | vedclass

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(B) To find the class mark $(x_{i})$ for each interval,the following relation is used:$x_{i} = \frac{	ext{Upper class limit} + 	ext{Lower class limi

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$20$

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(B) To find the class mark $(x_{i})$ for each interval,the following relation is used:$x_{i} = \frac{	ext{Upper class limit} + 	ext{Lower class limit}}{2}$Given that,mean pocket allowance,$\bar{x} = Rs. 18$.Taking $18$ as assumed mean $(a)$,$d_{i}$ and $f_{i}d_{i}$ are calculated as follows:Daily pocket allowance (in $Rs.$)Number of children $(f_{i})$Class mark $(x_{i})$$d_{i} = x_{i} - 18$$f_{i}d_{i}$$11-13$$7$$12$$-6$$-42$$13-15$$6$$14$$-4$$-24$$15-17$$9$$16$$-2$$-18$$17-19$$13$$18$$0$$0$$19-21$$f$$20$$2$$2f$$21-23$$5$$22$$4$$20$$23-25$$4$$24$$6$$24$Total$\sum f_{i} = 44 + f$$\sum f_{i}d_{i} = 2f - 40$From the table,we obtain:$\sum f_{i} = 44 + f$$\sum f_{i}d_{i} = 2f - 40$Using the formula for mean:$\bar{x} = a + \frac{\sum f_{i}d_{i}}{\sum f_{i}}$$18 = 18 + \left(\frac{2f - 40}{44 + f}ight)$$0 = \frac{2f - 40}{44 + f}$$2f - 40 = 0$$2f = 40$$f = 20$Hence,the missing frequency $f$ is $20$.

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 $7$ $6$ $9$ $13$ $f$ $5$ $4$

Similar Questions

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

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$