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(C) To find the class mark $(x_i)$ of each interval,the following formula is used:$x_i = \frac{	ext{Upper class limit} + 	ext{Lower class limit}}{2}

Vedclass · Accepted Answer

$75.9$

Vedclass · Answer

(C) To find the class mark $(x_i)$ of each interval,the following formula is used:$x_i = \frac{	ext{Upper class limit} + 	ext{Lower class limit}}{2}$The class size $(h)$ of this data is $3$.Taking $75.5$ as the assumed mean $(a)$,we calculate $d_i, u_i,$ and $f_i u_i$ as follows:Number of heartbeats per minuteNumber of women $(f_i)$$x_i$$d_i = x_i - 75.5$$u_i = \frac{d_i}{3}$$f_i u_i$$65$-$68$$2$$66.5$-$9$-$3$-$6$$68$-$71$$4$$69.5$-$6$-$2$-$8$$71$-$74$$3$$72.5$-$3$-$1$-$3$$74$-$77$$8$$75.5$$0$$0$$0$$77$-$80$$7$$78.5$$3$$1$$7$$80$-$83$$4$$81.5$$6$$2$$8$$83$-$86$$2$$84.5$$9$$3$$6$Total$30$---$4$From the table,we have $\sum f_i = 30$ and $\sum f_i u_i = 4$.The mean $\bar{x}$ is calculated using the step-deviation method:$\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} ight) 	imes h$$\bar{x} = 75.5 + \left( \frac{4}{30} ight) 	imes 3$$\bar{x} = 75.5 + 0.4 = 75.9$Therefore,the mean heartbeats per minute for these women is $75.9$.

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Number of heartbeats per minute	$65$-$68$	$68$-$71$	$71$-$74$	$74$-$77$	$77$-$80$	$80$-$83$	$83$-$86$
Number of women	$2$	$4$	$3$	$8$	$7$	$4$	$2$