Name: Variance and Standard Deviation for Ungrouped Distribution
Uploaded: 2019-07-19
Duration: 6 min 21 s
Description: In this video, let us learn to find the variance and standard deviation of ungrouped data. We will also learn the formula to find the discrete frequency dist...

Question

Calculate the mean and median for the following frequency distribution :

Wages (in ₹)	$15 - 19$	$20 - 24$	$25 - 29$	$30 - 34$	$35 - 39$	$40 - 44$	$45 - 49$	$50 - 54$	$55 - 59$	$60 - 64$
No. of workers	$31$	$47$	$59$	$78$	$104$	$113$	$81$	$60$	$52$	$25$

Also calculate $Q_{1}, Q_{3}, P_{10}, P_{90}$ .

Accepted Answer

Since the given distribution is discontinuous but equal class interval is of equal width. To make it continuous we use adjustment factor $= \frac{20 - 19}{2} = 0.5$

Now we subtract $0.5$ from lower limit and add $0.5$ to upper limit of each class intervals.

So we construct the following table of values is as under :

Classes Class mark $x_{i}$ $45 - 49$ Cumulative frequency $\begin{matrix} d_{i} = x_{i} - A \\ A = 37 \end{matrix}$ $31$ $f_{i} u_{i}$

$14.5 - 19.5 19.5 - 24.5 24.5 - 29.5 29.5 - 34.5 34.5 - 39.5 39.5 - 44.5 44.5 - 49.5 49.5 - 54.5 54.5 - 59.5 59.5 - 64.5$ $17 22 27 32 37 42 47 52 57 62$ $31 47 59 78 104 113 81 60 52 25$ $31 78 137 215 319 432 513 573 625 650$ $- 20 - 15 - 10 - 5 0 5 10 15 20 25$ $- 4 - 3 - 2 - 1 0 1 2 3 4 5$ $- 124 - 141 - 118 - 78 0 113 162 180 208 125$

$\sum f_{i} = 650$ $\sum f_{i} u_{i} = 327$

Thus by step deviation method,

Mean $\bar{x} = A + \frac{\sum f_{i} u_{i}}{\sum f_{i}} \times c = 37 + \frac{327}{650} \times 5 = 39.52$

Here $n = \sum f_{i} = 650 (even)$

$∴$ Median $= {(\frac{n}{2})}^{th} value = {(\frac{650}{2})}^{th} value = 325^{th} value$

Which lies in the class $39.5 - 44.5$

$l = 39.5; f = 113; c = 319; i = 5$

$∴$ Median $= l + \frac{\frac{n}{2} - c}{f} \times i = 39.5 + \frac{325 - 319}{113} \times 5 = 39.5 + \frac{30}{113} = 39.765$

$Q_{1} = {(\frac{n}{4})}^{th} value = {(\frac{650}{4})}^{th} value i . e 162 . 5^{th} value$

Which lies in the class $29.5 - 34.5$

Here $l = 29.5; f = 78; c = 137; i = 5$

$Q_{1} = l + \frac{\frac{n}{4} - c}{f} \times i = 29.5 + \frac{162.5 - 137}{78} \times 5 = 29.5 + 1.6346 = 31.13$

$Q_{3} = {(\frac{3 n}{4})}^{th} value = {(\frac{3 \times 650}{4})}^{th} value i . e 487 . 5^{th} value$

Which lies in the class $44.5 - 49.5$

Here $l = 44.5; f = 81; c = 432; i = 5$

$Q_{3} = l + \frac{\frac{3 n}{4} - c}{f} \times i = 44.5 + \frac{487.5 - 432}{81} \times 5 = 44.5 + 3.4259 = 47.93$

$P_{10} = {(\frac{10 n}{100})}^{th} value = \frac{10 \times 650}{100} th value i . e 65^{th} value$

Which lies in the class $19.5 - 24.5$

Here $l = 19.5; f = 47; c = 31; i = 5$

$P_{10} = l + \frac{\frac{10 n}{100} - c}{f} \times i = 19.5 + \frac{657.5 - 31}{47} \times 5 = 19.5 + 3.6170 = 23.1170$

$P_{90} = {(\frac{90 n}{100})}^{th} value = \frac{90 \times 650}{100} th value i . e 585^{th} value$

Which lies in the class $54.5 - 59.5$

Here, $l = 54.5; f = 52; c = 573; i = 5$

$P_{90} = l + \frac{\frac{90 n}{100} - c}{f} \times i = 54.5 + \frac{585 - 573}{52} \times 5 = 54.5 + 1.1538 = 55.65$

Hence, the required value of $Q_{1}, Q_{3}, P_{10}, P_{90}$ are $31.13, 47.93, 23.12, 55.65$

Marks obtained	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$
No. of students	$8$	$10$	$22$	$40$	$20$

More than (in ₹)	$240$	$210$	$180$	$150$	$120$	$90$	$60$
No. of workers	$0$	$4$	$27$	$39$	$50$	$57$	$60$

$x$	$14$	$15$	$16$	$17$	$18$	$19$	$20$
$f$	$14$	$26$	$18$	$9$	$2$	$1$	$1$

Score	$0$	$1$	$2$	$3$	$4$	$5$
Number of shots	$0$	$3$	$6$	$4$	$7$	$5$

Important Questions on Statistics

Learn and Prepare for any exam you want !

Calculate the mean and median for the following frequency distribution : Wages (in ₹) 15-19 20-24 25-29 30-34 35-39 40-44 45-49 50-54 55-59 60-64 No. of workers 31 47 59 78 104 113 81 60 52 25 Also calculate Q1, Q3, P10, P90.

Important Questions on Statistics

Learn and Prepare for any exam you want !