5 Marks Questions

Question 15 Marks

The heights (in cm) of $50$ students of a class are given below:

Height (in cm)	$156$	$154$	$155$	$151$	$157$	$152$	$153$
Number of students	$8$	$4$	$10$	$6$	$7$	$3$	$12$

Find the median height.

Answer

Arranging in order and then preparing its cumulative frequency table:

Height (in cm) (x)	Number of students (f)	c.f
$151$	$6$	$6$
$152$	$3$	$9$
$153$	$12$	$21$
$154$	$4$	$25$
$155$	$10$	$35$
$156$	$8$	$43$
$157$	$7$	$50$

Here, number of terms $(N) = 50$ which is even
$\text{Median}=\frac{1}{2}\Big\{\frac{1}{2}\text{th term}+\Big(\frac{\text{n}}{2}+1\Big)\text{term}\Big\}$
$=\frac{1}{2}\Big\{\frac{50}{2}\text{th term}+\Big(\frac{50}{2}+1\Big)\text{term}\Big\}$
$=\frac{1}{2}\{25\text{th term}+26\text{th term}\}$
$\frac{1}{2}\{154+155\}\\\ \{\therefore\ \text{Value of 25th}=154\ \text{value of}\ 26\ \text{to}\ 35=155\}$
$\frac{1}{2}\times309=154.5\text{cm}$
$\text{Hence median }=154.5\text{cm}$

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Question 25 Marks

The following table shows the marks obtained by $41$ students of a class.

Marks obtained	$15$	$17$	$20$	$22$	$25$	$30$
Number of students	$2$	$5$	$10$	$12$	$8$	$4$

Find the median and mean marks. Using empirical formula, calculate its mode.

Answer

We prepare the table as given below:

Marks obtained (x)	No. of students (f)	c.f	x × f
$15$	$2$	$2$	$30$
$17$	$5$	$7$	$85$
$20$	$10$	$17$	$200$
$22$	$12$	$29$	$264$
$25$	$8$	$37$	$200$
$30$	$4$	$41$	$120$
Total	$41$		$899$

$\text{Mean}=\frac{\sum\text{fx}}{\sum\text{f}}=\frac{899}{41}=21.92$
Here, number of terms $(N) = 41,$ which is odd
$\text{Median}=\frac{\text{n+1}}{2}\text{th term}=\frac{41+1}{2}\text{th term}$
$=\frac{42}{2}=21\text{th term}=22\{\text{value of 18 to 29=22}\}$
Mode $= 3($median$) - 2($mean$)$
$= 3 \times 22 - 2 \times 21.92 = 66 - 43.84 = 22.16$

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Question 35 Marks

The following table shows the weight of $12$ players:

Weight (in kg)	$48$	$50$	$52$	$54$	$58$
Number of players	$4$	$3$	$2$	$2$	$1$

Find the median and mean weights.
Using empirical formula, calculate its mode.

Answer

We prepare the table as given below:

Weight (in kg) (x)	Number of players (f)	c.f	x × f
$48$	$4$	$4$	$192$
$50$	$3$	$7$	$150$
$52$	$2$	$9$	$104$
$54$	$2$	$11$	$108$
$58$	$1$	$12$	$58$
Total	$12$		$612$

Number of terms $(N)$ is $12, $ which is an even number.
$\text{Median}=\frac{1}{2}\Big\{\Big(\frac{\text{N}}{2}\Big)\text{th observation}+\Big(\frac{\text{n}}{2}+1\Big)\text{th observation}\Big\}$
$=\{6\ \text{th observation}+7\text{th observation}\}$
$=\frac{1}{2}\{50+50\}$
$\text{Median}=50$
$\text{Mean}=\frac{\sum(\text{f}_\text{i}\times\text{x}_\text{i})}{\sum\text{f}_\text{i}}$
$=\frac{612}{12}$
Mean $= 51$
Using empirical formula:
Mode $= 3($Median$) - 2($Mean$)$
$= 150 - 102$
Mode $= 48$
Hence, the median is $50,$ the mean is $51$ and the mode is $48.$

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Question 45 Marks

Daily wages of $45$ workers in a factory are given below:

Daily wages (in Rs)	$300$	$375$	$450$	$525$	$600$
Number of workers	$6$	$8$	$9$	$12$	$10$

Find the median and the mean. Using empirical formula, calculate its mode.

Answer

We prepare the table as given below:

Daily wages (in Rs.) x	No. of workers f	c.f.	x \times f
$300$	$6$	$6$	$1800$
$375$	$8$	$14$	$3000$
$450$	$9$	$23$	$4050$
$525$	$12$	$35$	$6300$
$600$	$10$	$45$	$6000$
Total	$45$		$21150$

$\text{Mean}=\frac{\sum\text{fx}}{\sum\text{f}}=\frac{21150}{45}=470$
Here, number of terms $= 45,$
which is odd $\text{Median}=\frac{\text{n}+1}{2}\text{th term}=\frac{45+1}{2}=\frac{46}{2}\text{th term}$
$= 23th$ term $= 450$
Now, mode $= 3($median$) - 2($mean$) = 3 \times 450 - 2 \times 470$
$= 1350 - 940 = 410$
$$

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