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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 $75.9$.

Age (in years)	Number of policy holders
Below $20$	$2$
Below $25$	$6$
Below $30$	$24$
Below $35$	$45$
Below $40$	$78$
Below $45$	$89$
Below $50$	$92$
Below $55$	$98$
Below $60$	$100$

Length (in mm)	Number of leaves
$118-126$	$3$
$127-135$	$5$
$136-144$	$9$
$145-153$	$12$
$154-162$	$5$
$163-171$	$4$
$172-180$	$2$

Age (in years)	$5$-$15$	$15$-$25$	$25$-$35$	$35$-$45$	$45$-$55$	$55$-$65$
Number of patients	$6$	$11$	$21$	$23$	$14$	$5$

Family size	$1-3$	$3-5$	$5-7$	$7-9$	$9-11$
Number of families	$7$	$8$	$2$	$2$	$1$

Life time (in hours)	Number of lamps
$1500-2000$	$14$
$2000-2500$	$56$
$2500-3000$	$60$
$3000-3500$	$86$
$3500-4000$	$74$
$4000-4500$	$62$
$4500-5000$	$48$

Similar Questions

$A$ survey conducted on $20$ households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:
Family size $1-3$ $3-5$ $5-7$ $7-9$ $9-11$
Number of families $7$ $8$ $2$ $2$ $1$

Find the mode of this data.

The following table gives the distribution of the lifetime of $400$ neon lamps:
Life time (in hours) Number of lamps
$1500-2000$ $14$
$2000-2500$ $56$
$2500-3000$ $60$
$3000-3500$ $86$
$3500-4000$ $74$
$4000-4500$ $62$
$4500-5000$ $48$

Find the median lifetime of a lamp.

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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	$2$	$4$	$3$	$8$	$7$	$4$	$2$

Similar Questions

The lengths of $40$ leaves of a plant are measured correct to the nearest millimetre,and the data obtained is represented in the following table:Length (in mm)Number of leaves$118-126$$3$$127-135$$5$$136-144$$9$$145-153$$12$$154-162$$5$$163-171$$4$$172-180$$2$Find the median length of the leaves.

$A$ survey conducted on $20$ households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:Family size$1-3$$3-5$$5-7$$7-9$$9-11$Number of families$7$$8$$2$$2$$1$Find the mode of this data.

The following table gives the distribution of the lifetime of $400$ neon lamps:Life time (in hours)Number of lamps$1500-2000$$14$$2000-2500$$56$$2500-3000$$60$$3000-3500$$86$$3500-4000$$74$$4000-4500$$62$$4500-5000$$48$Find the median lifetime of a lamp.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ survey conducted on $20$ households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:
Family size $1-3$ $3-5$ $5-7$ $7-9$ $9-11$
Number of families $7$ $8$ $2$ $2$ $1$

Find the mode of this data.

The following table gives the distribution of the lifetime of $400$ neon lamps:
Life time (in hours) Number of lamps
$1500-2000$ $14$
$2000-2500$ $56$
$2500-3000$ $60$
$3000-3500$ $86$
$3500-4000$ $74$
$4000-4500$ $62$
$4500-5000$ $48$

Find the median lifetime of a lamp.