3 Marks Question

Question 513 Marks

Find the mean of each of the following frequency distributions:

Class interval	$25-35$	$35-45$	$45-55$	$55-65$	$65-75$
Frequency	$6$	$10$	$8$	$12$	$4$

Answer

Let the assume $(A) = 50$

Class interval	Mid$-$value $x_i$	$d_i = x_i - 50$	$\text{u}_\text{i}=\frac{\text{d}}{\text{h}}=\frac{\text{d}}{10}$	Frequency $f_i$	$f_iu_i$
$25-35$	$30$	$-20$	$-2$	$6$	$-12$
$35-45$	$40$	$-10$	$-1$	$10$	$-10$
$45-55$	$50$	$0$	$0$	$8$	$0$
$55-65$	$60$	$10$	$1$	$12$	$12$
$65-75$	$70$	$20$	$2$	$4$	$8$
				$N = 40$	$\sum\text{f}_\text{i}\text{u}_\text{i}=-2$

We have
$A = 50, h = 10$
$\text{Mean}=\text{A}+\text{h}\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\text{N}}$
$=50+10\Big(\frac{-2}{40}\Big)$
$=50-0.5$
$=49.5$

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Question 523 Marks

The arithmetic mean of the following data is 25, find the value of k.

x	5	15	25	35	45
f	3	k	3	6	2

Answer

x	f	fx
5	3	15
15	k	15k
25	3	75
35	6	210
45	2	90
Total	14 + k	390 + 15k

$\text{Mean}=\frac{\sum\text{f}\times\text{x}}{\sum\text{f}}$
$\Rightarrow25=\frac{390+15\text{k}}{14+\text{k}}$
$\Rightarrow25(14+\text{k})=390+15\text{k}$
$\Rightarrow350+25\text{k}=390+15\text{k}$
$\Rightarrow25\text{k}-15\text{k}=390-350$
$\Rightarrow10\text{k}=40$
$\Rightarrow\text{k}=\frac{40}{10}=4$
Hence, k = 4

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Question 533 Marks

The daily income of a sample of $50$ employees are tabulated as follows:

Income $($in $Rs.):$	$1-200$	$201-400$	$401-600$	$601-800$
No. of employees$:$	$14$	$15$	$14$	$7$

Find the mean daily income of employees.

Answer

Since, given data is not continuous, so we subtract $0.5$ from the lower limit and add $0.5$ in the upper limit of each class.
Now we first, find the class mark $xt$ of each class and then proceed as follows:

Income $($in $Rs.)$	Class marks $(x_i)$	Number of employees $(f_i)$	$\text{u}_\text{i}=\frac{\text{x}_\text{i}-\text{a}}{\text{h}}\\=\frac{\text{x}_\text{i}-300.5}{200}$	$f_iu_i$
$0.5-200.5$	$100.5$	$14$	$-1$	$-14$
$200.5-400.5$	$300.5$	$15$	$0$	$0$
$400.5-600.5$	$500.5$	$14$	$1$	$14$
$600.5-800.5$	$700.5$	$7$	$2$	$14$
		$\text{N}=\sum\text{f}_\text{i}=50$		$\sum\text{f}_\text{i}\text{u}_\text{i}=14$

$\therefore$ Assumed mean, $a = 300.5$
Class width, $h = 200$
and total observation, $N = 50$
By step deviation method,
$\text{Mean}=\text{A}+\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\times\text{h}$
$=300.5+\frac{14}{50}\times200$
$=300.5+14\times4$
$=300.5+56=356.5$

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Question 543 Marks

The annual rainfall record of a city for 66 days is given in the following tab:

Rainfull (in cm):	0-10	10-20	20-30	30-40	40-50	50-60
Number of days:	22	10	8	15	5	6

Calculate the median rainfall using ogives of more than type and less than type.

Answer

We observe that, the annual rainfall record of a city less than 0 is 0. Similarly, less than 10 include the annual rainfall record of a city from 0 as well as the annual rainfall record of a city from 0-10. So, the total annual rainfall record of a city for less than 10 cm is 0 + 22 = 22 days. Continuing in this manner, we will get remaining less than 20, 30, 40, 50 and 60. Also, we observe that annual rainfall record of a city for 66 days is more than or equal to 0 cm. Since, 22 days lies in the interval 0-10. So, annual rainfall record for 66 – 22 = 44 days is more than or equal to 10 cm. Continuing in this manner we will get remaining more than or equal to 20, 30 , 40, 50, and 60. Now, we construct a table for less than and more than type.

Less than type		More than type
Rainfull (in cm)	No. of days	Rainfull (in cm)	No. of days
Less than 0 Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60	0 0 + 22 = 22 22 + 10 = 32 32 + 8 = 40 40 + 15 = 55 55 + 5 = 60 60 + 6 = 66	More than or equal to 0 More than or equal to 10 More than or equal to 20 More than or equal to 30 More than or equal to 40 More than or equal to 50 More than or equal to 60	66 66 - 22 = 44 44 - 10 = 34 34 - 8 = 26 26 - 15 = 11 11 - 5 = 6 6 - 6 = 0

To draw less than type ogive we plot the points (0, 0), (10, 22), (20, 32), (30, 40), (40, 55), (50, 60), (60, 66) on the paper and join them by free hand. To draw the more than type ogive we plot the points (0, 66), (10, 44), (20, 34), (30, 26), (40, 11), (50, 6) and (60, 0) on the graph paper and join them by free hand.

Firstly, we plot a line parallel to X-axis at intersection point of both ogives, which further intersect at (0, 33) on Y- axis. Now, we draw a line perpendicular to X-axis at intersection point of both ogives, which further intersect at (21.25, 0) on X-axis. Which is the required median using ogives. Hence, median rainfall = 21.25 cm.

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Question 553 Marks

Find the value of $p$ for the following distribution whose mean is $16.6.$

$x$	$8$	$12$	$15$	$P$	$20$	$25$	$30$
$f$	$12$	$16$	$20$	$24$	$16$	$8$	$4$

Answer

Given:

$x_i$	$8$	$12$	$15$	$P$	$20$	$25$	$30$
$f_i$	$12$	$16$	$20$	$24$	$16$	$8$	$4$

Mean $= 16.6$
First of all prepare the frequency table in such a way that its first column consist of the values of the variate $(x_i)$ and the second column the corresponding frequencies$(f_i).$
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing$(f_{i }x_i).$
Then, sum of all entries in the column second and denoted by $\sum\text{f}_\text{i}$and in the third column to obtain$\sum\text{f}_\text{i}\text{x}_\text{i}$

$x_i$	$f_i$	$f_ix_i$
$8$	$12$	$96$
$12$	$16$	$192$
$15$	$20$	$300$
$P$	$24$	$24P$
$20$	$16$	$320$
$25$	$8$	$200$
$30$	$4$	$120$
	$\sum\text{f}_\text{i}=100$	$\sum\text{f}_\text{i}\text{x}_\text{i}=1228+24\text{P}$

We know that mean, $\overline{\text{X}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$16.6=\frac{1228+24\text{P}}{100}$
By using cross multiplication method
$1228+24\text{P}=16.6\times100$
$24\text{P}=1660-1228$
$=\frac{432}{24}$
$=18$
Hence,$\text{P}=18$

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Question 563 Marks

The. number of telephone calls received at an exchange per interval for $250$ successive one$-$minute intervals are given in the following frequency table$:$

No of calls $(x)$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
No of intervals $(f)$	$15$	$24$	$29$	$46$	$54$	$43$	$39$

Compute the mean number of calls per interval.

Answer

Let be assumed mean $(A) = 3$

No of calls $(x_i)$	No of intevals $(f_i)$	$u_{i }= x; - A = x_i - 3$	$f_iu_i;$
$0$	$15$	$-3$	$-45$
$1$	$24$	$-2$	$-48$
$2$	$29$	$-1$	$-29$
$3$	$46$	$0$	$0$
$4$	$54$	$1$	$54$
$5$	$43$	$2$	$86$
$6$	$39$	$3$	$117$
	$N = 250$		$\sum\text{f}_\text{i}\text{u}_\text{i}=135$

Mean number of calls $=\text{A}+\frac{\sum\text{f}_\text{i}\text{u}_\text{i};}{\text{N}}$
$=3+\frac{135}{250}$
$=\frac{750+135}{250}$
$=\frac{885}{250}$
$=3.54$

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Question 573 Marks

The following table gives the height of trees:

Height	No. of trees
Less than 7	26
Less than 14	57
Less than 21	92
Less than 28	134
Less than 35	216
Less than 42	287
Less than 49	341
Less than 56	360

Draw 'less than' ogive and 'more than' ogive.

Answer

Consider the following table.

Height (less than)	Height-class	No. of trees	Cumulative frequency	Suitable points
7	0-7	26	26	(7, 26)
14	7-14	31	57	(14, 57)
21	14-21	35	92	(21, 92)
28	21-28	42	134	(28, 134)
35	28-35	82	216	(35, 216)
42	35-42	71	287	(42, 287)
49	42-49	54	341	(49, 341)
56	49-56	19	360	(56, 360)

Now, draw the less than ogive using suitable points.

Now, prepare the cumulative frequency table for more than series.

Heith	No. of trees	Height (more than)	Cumulative frequency	Suitable points
0-7	26	0	360	(0, 360)
7-14	31	7	334	(7, 334)
14-21	35	14	303	(14, 303)
21-28	42	21	268	(21, 268)
28-35	82	28	226	(28, 226)
35-42	71	35	144	(35, 144)
42-49	54	42	73	(42, 73)
49-56	19	49	54	(49, 54)

Now, draw the more than ogive using suitable points.

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Question 583 Marks

The following distribution shows the daily pocket allowance given to the children of a multistory building. The average pocket allowance is $Rs. 18.00.$ Find out the missing frequency.

Class interval	$11-13$	$13-15$	$15-17$	$17-19$	$19-21$	$21-23$	$23-25$
Frequency	$7$	$6$	$9$	$13$	$-$	$5$	$4$

Answer

Given mean $= 18,$ let missing frequency be $v$

Class interval	Mid$-$value	Frequency $ f_i$	$f_iu_i$
$11-13$	$12$	$7$	$83$
$13-15$	$14$	$6$	$88$
$15-17$	$16$	$9$	$144$
$17-19$	$18$	$9$	$234$
$19-21$	$20$	$x$	$20x$
$21-23$	$22$	$5$	$110$
$23-25$	$14$	$4$	$56$
		$N = 44 + x$	$752 + 20x$

Mean $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}$
$18=\frac{752+20\text{x}}{44+\text{x}}$
$792+18\text{x}=752+20\text{x}$
$2\text{x}=40$
$\text{x}=20$

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Question 593 Marks

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

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

Find the mean of each of the following frequency distribution $ (5 - 14)$

Answer

We may find class marks of each interval $(x_i)$ by using the relation
$\text{x}_\text{i}=\frac{\text{Upper class limit+lower limit class limit}}2{}$
Class size of this data $= 3$
Now taking $75.5$ as assured mean $(a)$ we
May calculate,$ d_{i, }u_i, f, u,$ as following.

Number of heart beat per minnute	Number of women$(x_i)$	$x_i$	$d_i = x_i - 75.5$	$\text{u}_\text{i}=\frac{\text{x}_\text{i}-755}{\text{h}}$	$f_iu_i$
$65-68$	$2$	$66.5$	$-9$	$-3$	$-6$
$68-71$	$4$	$69.5$	$-6$	$-2$	$-8$
$71-74$	$3$	$72.5$	$-3$	$-1$	$-3$
$74-77$	$8$	$75.5$	$0$	$0$	$0$
$77-80$	$7$	$78.5$	$3$	$1$	$7$
$80-83$	$4$	$81.5$	$6$	$2$	$8$
$83-86$	$2$	$84.5$	$9$	$3$	$6$
	$30$				$4$

Now we may observe from table that $ \sum\text{f}_\text{i}= 30,\sum\text{f}_\text{i}\text{u}_\text{i}=4$
Mean $\overline{\text{x}}=\text{A}+\text{h}\Big[\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\Big]=75.5+3\Big(\frac{4}{30}\Big)$
$=75.5+0.4=75.9$
So mean hear beats per minute forn those women are $75.9$ beats per minute.

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Question 603 Marks

The following distribution gives the daily income of 50 workers of a factory:

Daily income (in ₹)	100-120	120-140	140-160	160-180	180-200
Number of workers	12	14	8	6	10

Convert the above distribution to a less than type cumulative frequency distribution and draw its ogive.

Answer

We first prepare the cumulative frequency table by less thsn method as given below.

Daily income (in ₹)	Cumulative frequency
<120	12
<140	26
<160	34
<180	40
<200	50

Now, we mark on x– axis upper class limit, y– axis cumulative frequencies.
Thus, we plot the points (120, 12) (140, 26) (160, 34) (180, 40) (200, 50)

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Question 613 Marks

Find the mean of each the following frequency distributions: $(5 – 14)$

Class interval	$0-6$	$6-12$	$12-18$	$18-24$	$24-30$
Frequency	$6$	$8$	$10$	$9$	$7$

Answer

Let a assume mean be $15$

Class interval	Mid$-$value $x_i$	$d_i = x_i - 15$	$\text{u}_\text{i}=\frac{\text{x}_\text{i}-15}{6}$	$f_i$	$f_iu_i$
$0-6$	$3$	$-12$	$-2$	$6$	$-12$
$6-12$	$9$	$-6$	$-1$	$8$	$-8$
$12-18$	$15$	$0$	$0$	$10$	$0$
$18-24$	$21$	$6$	$1$	$9$	$9$
$24-30$	$27$	$18$	$2$	$7$	$14$
Total				$N = 40$	$3$

$A = 15, h = 6$
$\text{Mean}=\text{A}+\text{h}\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}$
$=15+6\times\frac{3}{40}$
$= 15 + 0.45$
$= 15.45$

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Question 623 Marks

Calculate the median from the following data:

Rent (in Rs.)	15-25	25-35	35-45	45-55	55-65	65-75	75-85	85-95
No. of Houses	8	10	15	25	40	20	15	7

Answer

Writing the given data in cumulative frequency table as shown

Rent (in Rs.)	No. of houses (f)	e.f.
15-25	8	8
25-35	10	18
35-45	15	33
45-55	25	58
55-65	40	98
65-75	20	118
75-85	15	133
85-95	7	140

Here N = 140
$\therefore \frac{\text{N}}{2}=\frac{140}{2}=70$
$\therefore$ Corresponding mediam class $=55-65$
$\text{l}=55,\text{f}=40,\text{F}=58$ and h = 10
$\therefore \text{mediam}=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=55+\frac{70-58}{40}\times10$
$\Rightarrow 55+\frac{12}{40}\times10$
$=55+3=58$

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Question 633 Marks

Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:
715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719

Answer

Arranging in ascending order, we get 694, 696, 699, 705, 710, 712, 715, 716, 719, 724, 725, 728, 729, 734, 745Here N = 15 which is odd
Median $=\frac{\text{N}+1}{2}$ th term $=\frac{15+1}{2}$ th term = 8th term Which is 716 Hence median = 716

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Question 643 Marks

The arithmetic mean of the following data is $14,$ find the value of $k.$

$x$	$5$	$10$	$15$	$20$	$25$
$f$	$7$	$k$	$8$	$4$	$5$

Answer

Given:

$x_i$	$5$	$10$	$15$	$20$	$25$
$f_i$	$7$	$k$	$8$	$4$	$5$

Mean $= 14$
First of all prepare the frequency table in such a way that its first column consist of the values of the variate$(x_i)$ and the second column the corresponding frequencies$(f_i).$
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing$(f_ix_i).$
Then, sum of all entries in the column second and denoted by $\sum\text{f}_\text{i}$ and in the third column to obtain $\sum\text{f}_\text{i}\text{x}_\text{i}$.

$x_i$	$f_i$	$f_ix_i$
$5$	$7$	$35$
$10$	$k$	$10k$
$15$	$8$	$120$
$20$	$4$	$80$
$25$	$5$	$125$
	$\sum\text{f}_\text{i}=24+\text{k}$	$\sum\text{f}_\text{i}\text{x}_\text{i}=360+10\text{k}$

We know that mean, $\overline{\text{X}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$14=\frac{360+10\text{k}}{24+\text{k}}$
By using cross multiplication method,
$336+14\text{k}=360+10\text{k}$
$14\text{k}-10\text{k}=360-336$
$4\text{k}=24$
$\text{k}=6$
Hence, $k = 6$

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Question 653 Marks

Find the mean of each of the following frequency distributions$:$

Class interval	$0-8$	$8-16$	$16-24$	$24-32$	$32-40$
Frequency	$6$	$7$	$10$	$8$	$9$

Answer

Let assured mean $(A) = 20$

Class inteval	Class Marks $(x)$	Requency $(f)$	$d = x - A A = 20$	$f_id_i$
$0-8$	$4$	$6$	$-16$	$-96$
$8-16$	$12$	$7$	$-8$	$-56$
$16-24$	$20 - A$	$10$	$0$	$0$
$24-32$	$28$	$8$	$8$	$64$
$32-40$	$36$	$9$	$16$	$144$
Total		$40$		$56$

Mean $\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}} =20+\frac{56}{40}=20+\frac{7}{5}$
$= 20 + 1.4 = 21.4$

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Question 663 Marks

The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution:

Profit (in lakhs in ₹)	No. of shops (frequency)
More than or equal to 5	30
More than or equal to 10	28
More than or equal to 15	16
More than or equal to 20	14
More than or equal to 25	10
More than or equal to 30	7
More than or equal to 35	3

Draw both ogives for the above data and hence obtain the median.

Answer

Profit (in lakhs in ₹)	Class intervals	Number of shops (c.f.)	Frequency
More than or equal to 5	5-10	30	2
More than or equal to 10	10-15	28	12
More than or equal to 15	15-20	16	2
More than or equal to 20	20-25	14	4
More than or equal to 25	25-30	10	3
More than or equal to 30	30-35	7	4
More than or equal to 35	35-40	3	3

Now plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) on the graph and join them to get a more than curve.
Less than curve:

Class intervals	Frequency	c.f.
5-10	3	3
10-15	4	7
15-20	3	10
20-25	4	14
25-30	2	16
30-35	12	28
35-40	2	30

Now plot the points (10, 3), (15, 7), (20, 10), (25, 14), (30, 16), (35, 28) and (40, 30) on the graph and join them to get a less them ogive. The two curved intersect at P. From P, draw PM 1 x-axis, M is the median which is 22.5
$\therefore$ Median = Rs. 22.5 lakh

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Question 673 Marks

Draw an ogive by less than method for the following data:

No. of rooms	1	2	3	4	5	6	7	8	9	10
No. of houses	4	9	22	28	24	12	78	6	5	2

Answer

No. of rooms	N. of houses (f)	c.f.
1	4	4
2	9	13
3	22	35
4	28	63
5	24	87
6	12	99
7	8	107
8	6	113
9	5	118
10	2	120

Take numbers of rooms along the x-axis and c.f along the y-axis Plot the points (1, 4), (2, 13), (3, 35), (4, 63), (5, 87), (6, 99), (7, 107), (8, 113), (9, 118) and (10, 120) on the graph and join them and with free hand to get an ogive as shown. This is the less than ogive.

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Question 683 Marks

For the following distribution, calculate mean using all suitable methods.

Size of item	$1-4$	$4-9$	$9-16$	$16-27$
Frequency	$6$	$12$	$26$	$20$

Answer

Let assumed mean $(A) = 12.5$

Size of item	Class Marks $(x)$	Frequency $(f)$	$d_i = x - A A = 12.5$	$f_id_i$
$1-4$	$2.5$	$6$	$-10$	$-60$
$4-9$	$6.5$	$12$	$-6$	$-72$
$9-16$	$12.5 - A$	$26$	$0$	$0$
$16-27$	$21.5$	$20$	$9$	$180$
Total		$64$		$48$

$\text{Mean}=\text{A}+\text{hx}\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}=12.5+\frac{48}{64}$
$=12.5+\frac{3}{4}=12.5+0.75=13.25$

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Question 693 Marks

The monthly income of $100$ families are given as below:

Income in ₹	Number of families
$0-5000$	$8$
$5000-10000$	$26$
$10000-15000$	$41$
$15000-20000$	$16$
$20000-25000$	$3$
$25000-30000$	$3$
$30000-35000$	$2$
$35000-40000$	$1$

Calculate the modal income.

Answer

In a given data, the highest frequency is $41,$ which lies in the interval $10000-15000.$
Here, $l = 10000, f_1 = 41, f_0 = 26, f_2 = 16$ and $h = 5000$
$\therefore\text{Modal}=\text{l}+\Big(\frac{\text{f}_1-\text{f}_0}{\text{2f}_1-\text{f}_0-\text{f}_2}\Big)\times\text{h}$
$=10000+\Big(\frac{41-26}{2\times41-26-16}\Big)\times5000$
$=10000+\Big(\frac{15}{82-42}\Big)\times5000$
$=10000+\Big(\frac{15}{40}\Big)\times5000$
$=10000+15\times125=10000+1875=₹11875$
Hence, the modal income is $₹ \ 11875.$

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Question 703 Marks

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of a plants in $20$ houses in a locality. Find the mean number of plants per house.

Number of plants	$0-2$	$2-4$	$4-6$	$6-8$	$8-10$	$10-12$	$12-14$
Number of Houses	$1$	$2$	$1$	$5$	$6$	$2$	$3$

Which method did you use for finding the mean, and why?

Answer

We may prepare the table as shown:

No. of plants	Mid value $(x_i)$	No. of Houses $(f_i)$	$f_ix_i$
$0-2$	$1$	$1$	$1$
$2-4$	$3$	$2$	$6$
$4-6$	$5$	$1$	$5$
$6-8$	$7$	$5$	$35$
$8-10$	$9$	$6$	$54$
$10-12$	$11$	$2$	$22$
$12-14$	$13$	$3$	$39$
		$\sum\text{f}_\text{i}=20$	$\sum\text{f}_\text{i}\text{x}_\text{i}=162$

We know that mean, $\overline{\text{X}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$=\frac{162}{20}$
$=8.1$
Hence, mean $=8.1$
Direct method is easier than other methods. Therefore, we used direct method.

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Question 713 Marks

The following is the cummulative frequency distribution $($of less than type$)$ of $1000$ persons each of age $20$ years and above. Determine the mean age.

Age below $($in year$):$	$30$	$40$	$50$	$60$	$70$
Number of persons$:$	$100$	$220$	$350$	$750$	$950$

Answer

First, we make the frequency distribution of the given data and then proceed to calculate mean by computing class marks $(x_i), u_i $ and $f_iu_i$ as follows:

Class	Frequency $(f_i)$	Class mark $(x_i)$	$\text{u}_\text{i}=\frac{\text{x}_\text{i}-45}{10}$	$f_iu_i$
$20-30$	$100$	$25$	$-2$	$-200$
$30-40$	$120$	$35$	$-1$	$-120$
$40-50$	$130$	$45$	$0$	$0$
$50-60$	$400$	$55$	$1$	$400$
$60-70$	$200$	$65$	$2$	$400$
$70-80$	$50$	$75$	$3$	$150$
	$\sum\text{f}_\text{i}=1000$			$\sum\text{f}_\text{i}\text{u}_\text{i}=630$

We have taken assumed mean $(a) = 45. $Here, $h =$ class size $= 10$ using the formula
$\text{Mean}\ \overline{\text{x}}=\text{a}+\text{h}\Big(\frac{{{\sum\text{f}_\text{i}\text{u}_\text{i}}}}{{\sum\text{f}_\text{i}}}\Big)$
$=45+10\Big(\frac{630}{1000}\Big)$
$=45+6.3=51.3$
Thus, the mean age is $51.3$ years.

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Question 723 Marks

Find the missing value of p for the following distribution whose mean is 12.58.

x	5	8	10	12	P	20	25
f	2	5	8	22	7	4	2

Answer

Mean = 12.58

x	f	fx
5	2	10
8	5	40
10	8	80
12	22	264
P	7	7P
20	4	80
25	2	50
Total	50	524 + 7P

Mean$=\frac{\sum\text{f}\text{x}}{\sum\text{f}}$
$\Rightarrow12.58=\frac{524+7\text{P}}{50}$
$\Rightarrow524+7\text{P}=50\times12.58=629.00$
$\Rightarrow\text{P}=\frac{105}{7}=15$
$\therefore\text{P}=15$

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Question 733 Marks

During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg)	No. of students
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula

Answer

Weight (in kg)	No. of students	Class	c.f.
Less than 38	0	36-38	0
Less than 40	3	38-40	3
Less than 42	5	40-42	2
Less than 44	9	42-44	4
Less than 46	14	44-46	5
Less than 48	28	46-48	14
Less than 50	32	48-50	4
Less than 52	35	50-52	3

Plot the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35) on the graph and join them in free hand to get an ogive as shown.
Here N = 35 which is odd
$\therefore\frac{\text{N}}{2}=\frac{25}{2}=17.5$
From 17.5 on y-axis draw a line parallel to x-axis meeting the curve at P.
From P, draw PM ⊥ x-axis
$\therefore$ Median which is 46.5 (approx)
Now N = 17.5 lies in the class 46 – 48 (as 14 < 17.5 < 28)
$\therefore$ 46-48 is the median class
$\therefore\text{Median}=\text{I}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=46+\frac{17.5-14}{14}\times2$
$=46+\frac{3.5\times2}{14}$
$=46+0.5=46.5$

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Question 743 Marks

Consider the following distribution of daily wages of $50$ workers of a facotry.

Daily wages $($in $Rs.)$	$100-120$	$120-140$	$140-160$	$160-180$	$180-200$
Number of workers	$12$	$14$	$8$	$6$	$10$

Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer

Let assumed mean $(A) = 150$

Daily wages $($in $Rs.)$	Class Marks $(x)$	No. of workers $(f)$	$di = x_{i }- A A = 150$	$\text{u}_\text{i}=\frac{\text{d}}{\text{h}}$ $\text{h}=20$	$u_{i }f_i$
$100-120$	$110$	$12$	$-40$	$-2$	$-24$
$120-140$	$130$	$14$	$-20$	$-1$	$-14$
$140-160$	$150 - A$	$8$	$0$	$0$	$0$
$160-180$	$170$	$6$	$20$	$1$	$6$
$180-200$	$190$	$10$	$40$	$2$	$20$
Total		$50$			$-12$

Mean $=\text{A}+\text{h}\times \frac{\sum\text{u}_\text{i}\text{f}_\text{i}}{\sum\text{f}_\text{i}}$
$= 150+20\times\frac{-12}{50}=150-\frac{240}{50}$
$=150-4.80=145.20$
Mean daily wages per worker $= Rs. 145.20$

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Question 753 Marks

The following data gives the information on the observed lifetimes $($in hours$)$ of $225$ electrical components:

Lifetimes (in hours)	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	$100-120$
No. of components	$10$	$35$	$52$	$61$	$38$	$29$

Determine the modal lifetimes of the components.

Answer

Lifetimes $($in hours$)$	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	$100-120$
No. of components	$10$	$35$	$52$	$61$	$38$	$29$

Here, the maxiimum frequency of electrical components is $61$ so the modal class is $60-80.$
therefore,
$l = 60$
$h = 20$
$f = 61$
$f1 = 52$
$f2 = 38$
$\Rightarrow\text{Mode}=\text{l}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=60+\frac{61-52}{122-52-38}\times20$
$=60+\frac{9}{32}\times20$
$\text{Mode}=65.25\text{hrs}$
Thus the lifetimes of the componets is $65.625$ hours.

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Question 763 Marks

The following is the distribution of height of students of a certain class in a certain city:

Hight $($in $cms)$	$160-162$	$163-165$	$166-168$	$169-171$	$172-174$
No. of students	$15$	$118$	$142$	$127$	$18$

Answer

Hight $($in $cms)$	$160-162$	$163-165$	$166-168$	$169-171$	$172-174$
No. of students	$15$	$118$	$142$	$127$	$18$

Here model class is $165.5$
and $l = 165.5 h = 3, f = 142, f_{1 }= 118, f_{2 }= 127$
$\therefore\text{Mode}=\text{l}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=165.5+\frac{142-118}{2\times142-118-127}\times3$
$=165.5+\frac{24\times3}{284-245}$
$=165.5+\frac{72}{39}$
$=165.5+1.8=167.3$

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Question 773 Marks

The following table gives the number of branches and number of plants in the garden of a school.

No of branches $(x)$	$2$	$3$	$4$	$5$	$6$
No of plants $(x)$	$49$	$43$	$57$	$38$	$13$

Calculate the average number of branches per plant.

Answer

Let assumed mean $(A) = 4$

No of brances $(x_i)$	No of plants $n (f_i)$	$d_i = x_i - A, A = 4$	$f_i\times d_i$
$2$	$A49$	$-2$	$-98$
$3$	$43$	$-1$	$-43$
$4$	$57$	$0$	$0$
$5$	$38$	$1$	$38$
$6$	$13$	$2$	$26$
Total	$200$	$ $	$-77$

$\text{Mean}=\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}=4+\frac{-77}{200}=4-0.385$
$=3.615=3.62(\text{approx})$
$\therefore$ Mean number of branches per plant $= 3.62$

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Question 783 Marks

The following distribution gives the number of accidents met by $160$ workers in a factory during a month,

No. of accidents $(x)$	$0$	$1$	$2$	$3$	$4$
No. of workera $(f)$	$70$	$52$	$34$	$3$	$1$

Find the average number of accidents per worker.

Answer

Let the assume mean be $A = 2.$

No. of accidents $(x_i)$	No. of workers $(f_i)$	$d_i = x_i - A = x_i - 2$	$f_id_i$
$0$	$70$	$-2$	$-140$
$1$	$52$	$-1$	$-52$
$2$	$34$	$0$	$0$
$3$	$3$	$1$	$3$
$4$	$1$	$2$	$2$
	$\sum\text{f}_\text{i}=160$		$\sum\text{f}_\text{i}\text{d}_\text{i}=-187$

We know that mean, $\overline{\text{x}}=\text{A}+\frac{1}{\text{N}}\sum\limits_\text{i+1}^\text{n}\text{f}_\text{i}\text{d}_\text{i}$
Now, we have $\text{N}=\sum\text{f}_\text{i}=160,\sum\text{f}_\text{i}\text{d}_\text{i}=-187$ and $\text{A}=2$
Putting the values in the above formula, we get
$\overline{\text{x}}=\text{A}+\frac{1}{\text{N}}\sum\limits_\text{i+1}^\text{n}\text{f}_\text{i}\text{d}_\text{i}$
$=2+\frac{1}{160}\times (-187)$
$=2-\frac{187}{160}$
$=2-1.168$
$=0.83$
Hence, the average number of accidents per worker is $0.83.$

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Question 793 Marks

Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss:

No of head per toss	No. of tosses
0	38
1	144
2	342
3	287
4	164
5	25
Total	1000

Answer

No. of head per toss	No. of tosses
0	38
1	144
2	342
3	287
4	164
5	25

No. of heads per toss	No. of tosses	fx
0	28	0
1	144	144
2	342	684
3	287	861
4	164	656
5	25	125

Mean number of head per toss $=\frac{\sum\text{fx}}{\text{N}}$
$=\frac{2470}{1000}$
$=2.47$
$\text{Mean}=2.47$

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Question 803 Marks

A student noted the number of cars passing through a spot on a road for $100$ periods each of $3$ minutes and summarised it in the table given below. Find the mode of the data.

Number of cars	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70--80$
Frequency	$7$	$14$	$13$	$12$	$20$	$11$	$15$	$8$

Answer

No. of cars	Frequency
$0-10$	$7$
$10-20$	$14$
$20-30$	$13$
$30-40$	$12$
$40-50$	$20$
$50-60$	$11$
$60-70$	$15$
$70-80$	$8$

We see that class $ 40-50$ has the maximum frequency
$\therefore$ lt is a modal class
Here $l = 40, f = 20, f_1 = 12, f_2 = 11, h = 10$
$\text{Mode}=\text{l}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=40+\frac{20-12}{2\times20-12-11}\times10$
$=40+\frac{8\times10}{40-23}=40+\frac{8\times10}{17}$
$=40+\frac{80}{17}=40+4.70=44.7\text{ cars}$

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Question 813 Marks

Find the mean of the following data:

$x$	$19$	$21$	$23$	$25$	$27$	$29$	$31$
$f$	$13$	$15$	$16$	$18$	$16$	$15$	$13$

Answer

Given:

$x_1$	$19$	$21$	$23$	$25$	$27$	$29$	$31$
$f_1$	$13$	$15$	$16$	$18$	$16$	$15$	$13$

First of all prepare the frequency table in such a way that its first column consist of the values of the variate$(x_1)$ and the second column the corresponding frequencies$(f_1).$
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing$(f_1x_1).$
Then, sum of all entries in the column second and denoted by $\sum\text{f}_1$ and in the third column to obtain$\sum\text{f}_1\text{x}_1$.

$\text{x}_1$	$\text{f}_1$	$\text{f}_1\text{x}_1$
$19$	$13$	$247$
$21$	$15$	$315$
$23$	$16$	$368$
$25$	$18$	$450$
$27$	$16$	$432$
$29$	$15$	$435$
$31$	$13$	$403$
	$\sum\text{f}_1=106$	$\sum\text{f}_1\text{x}_1=2650$

We know that mean, $\overline{\text{x}}=\frac{\sum\text{f}_1\text{x}_1}{\sum\text{f}_1}$
$=\frac{2650}{106}$
$= 25$
Hence, mean $= 25$

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Question 823 Marks

The following table gives the daily income of $50$ workers of a factory:

Daily income $($in $₹)$	$100-120$	$120-140$	$140-160$	$160-180$	$180-200$
Number of workers	$12$	$14$	$8$	$6$	$1$

Find the mean, mode and median of the above data.

Answer

Daily income $($in $₹)$	No. of workers $f_1$	Mid value $x_1$	$\text{d}_1=\frac{\text{x}_1-150}{20}$	$f.d.$
$100-120$	$12$	$110$	$-2$	$-24$
$120-140$	$14$	$130$	$-1$	$-14$
$140-160$	$8$	$a = 150$	$0$	$0$
$160-180$	$6$	$170$	$1$	$6$
$180-200$	$10$	$190$	$2$	$20$
Total	$50$			$-12$

Here $i = 20$ and $AM = 150$
$\text{Mean}=\text{a}+\frac{\sum\text{f}\text{d}_i}{\sum\text{f}_i}\times\text{i}$$=150+\frac{-12}{50}\times20=150-4.8=145.2$
Mix frequency $= 14$
$\therefore\text{Modle Class}=120-140$
Now $\text{Mode}=\text{I}+\frac{\text{f}_1-\text{f}_2}{2\text{f}_1-\text{f}_0-\text{f}_2}$
$=120+\frac{2}{28-20}\times20=120+\frac{40}{8}$
$=120+5=125$$\text{n}=5$
$\therefore\text{Median class}=120-140$$\therefore\text{Medim}=\text{l}+\bigg(\frac{\frac{\text{n}}{2}-\text{c.f.}}{\text{f}}\bigg)\times\text{i}$
$=120+\frac{25-12}{14}\times20=120+\frac{13\times20}{14}=120+\frac{130}{7}$
$=120+18.57=138.57$
$\therefore\text{Median}=138.57$

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Question 833 Marks

The following table gives the number of boys of a particular age in a class of $40$ students. Calculate the mean age of the students:

Age$($in years$)$	$15$	$16$	$17$	$18$	$19$	$20$
No. of students	$3$	$8$	$10$	$10$	$5$	$4$

Answer

Given,

Age $($in years$): x_i$	$15$	$16$	$17$	$18$	$19$	$20$
No.of students$: f_i$	$3$	$8$	$10$	$10$	$5$	$4$

First of all prepare the frequency table in such a way that its first column consist of the values of the variate$(x_i)$ and the second column the corresponding frequencies$(f_i).$
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing$(f_ix_i).$
Then, sum of all entries in the column second and denoted by $\sum\text{f}_\text{i}$ and in the third column to obtain $\sum\text{f}_\text{i}\text{x}_\text{i}$

Age $($in years$) x_i$	No. of students $f_i$	$f_ix_i$
$15$	$3$	$45$
$16$	$8$	$128$
$17$	$10$	$170$
$18$	$10$	$180$
$19$	$5$	$95$
$20$	$4$	$80$
	$\sum\text{f}_\text{i}=40$	$\sum\text{f}_\text{i}\text{x}_\text{i}=698$

We know that mean, $\overline{\text{X}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\overline{\text{X}}=\frac{698}{40}$
$=17.45$
Hence, the mean age of the students $=17.45\ \text{years}$

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Question 843 Marks

Find the mean of each of the following frequency distributions:

Class interval	$0-8$	$8-16$	$16-24$	$24-32$	$32-40$
Frequency	$5$	$9$	$10$	$8$	$8$

Answer

Let the assumed mean $(A) = 20$

Class interval	Mid$-$value	$d_i = x_{i }- 20$	$\text{u}_\text{i}=\frac{\text{d}}{\text{h}}=\frac{\text{d}}{8}$	Frequency $f_i$	$f_iu_i$
$0-8$	$4$	$-16$	$-2$	$5$	$-10$
$8-16$	$12$	$-8$	$-1$	$9$	$-9$
$16-24$	$20$	$0$	$0$	$10$	$0$
$24-32$	$28$	$8$	$1$	$8$	$8$
$32-40$	$36$	$16$	$2$	$8$	$16$
				$N = 40$	$\sum\text{f}_\text{i}\text{u}_\text{i}=5$

We have
$\text{A}=20,\text{h}-8$
$\text{Mean}=\text{A}+\text{h}\times\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\text{N}}$
$=20+8\times\frac{5}{40}$
$=20+1$
$=21$

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Question 853 Marks

The marks obtained out of $50,$ by $102$ students in a Physics test are given in the frequency table below:

Marks $(x)$	$15$	$20$	$22$	$24$	$25$	$30$	$33$	$38$	$45$
Frequncy $(f)$	$5$	$8$	$11$	$20$	$23$	$18$	$13$	$3$	$1$

Find the average number of marks.

Answer

Let the assume mean be $A = 25$

Marker $(x_i):$	Frequency $(f_i):$	$d_i = x_i - A = x_{i }- 25$	$f_id_i$
$15$	$5$	$-10$	$-50$
$20$	$8$	$-5$	$-40$
$22$	$11$	$-3$	$-33$
$24$	$20$	$-1$	$-20$
$25$	$23$	$0$	$0$
$30$	$18$	$5$	$90$
$33$	$13$	$8$	$104$
$38$	$3$	$13$	$39$
$45$	$1$	$20$	$20$
	$\sum\text{f}_\text{i}=102$		$\sum\text{f}_\text{i}\text{d}_\text{i}=110$

We know that mean, $\overline{\text{x}}=\text{A}+\frac{1}{\text{N}}\sum\limits_\text{i+1}^\text{n}\text{f}_\text{i}\text{d}_\text{i}$
Now, we have $\text{N}=\sum\text{f}_\text{i}=102,\sum \text{f}_\text{i}\text{d}_\text{i}=110$ and $A = 25$
Putting the values in the above formula, we ge
$\overline{\text{x}}=\text{A}+\frac{1}{\text{N}}\sum\limits_\text{i+1}^\text{n}\text{f}_\text{i}\text{d}_\text{i}$
$=25+\frac{1}{102}\times(110)$
$=25+\frac{110}{102}$
$=25+1.078$
$=26.078$
$=26.08\ (\text{approximate})$
Hence, the average number of marks is $26.08.$

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Question 863 Marks

Five coins were simultaneously tossed $1000$ times, and at each toss the number of heads was observed. The number of tosses during which $0, 1, 2, 3, 4$, and $5$ heads were obtained are shown in the table below. Find the mean number of heads per toss:

of heads per toss $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
No of tosses $ (f)$	$38$	$144$	$342$	$287$	$164$	$25$

Answer

Let the assume mean be $A = 2.$

No. of heads per toss $x_i:$	No. of toss $f_i:$	$d_i = x_i - A = x_i - 2$	$f_id_i$
$0$	$38$	$-2$	$-76$
$1$	$144$	$-1$	$-144$
$2$	$342$	$0$	$0$
$3$	$287$	$1$	$287$
$4$	$164$	$2$	$328$
$5$	$25$	$3$	$75$
	$\sum\text{f}_\text{i}=1000$		$\sum\text{f}_\text{i}\text{d}_\text{i}=470$

We kniw that mean $\overline{\text{x}}=\text{A}+\frac{1}{\text{N}}\sum\limits_\text{i+1}^\text{n}\text{f}_\text{i}\text{d}_\text{i}$
Now, we have $\text{N}=\sum\text{f}_\text{i}=1000, \sum\text{f}_\text{i}\text{d}_\text{i}=470 $ and $A = 2$
Putting the values above in formula, we have
$\overline{\text{x}}=\text{A}+\frac{1}{\text{N}}\sum\limits_\text{i+1}^\text{n}\text{f}_\text{i}\text{d}_\text{i}$
$=2+\frac{1}{1000}\times470$
$=2+0.47$
$=2.47$
Hence, the mean number of heads per toss is $2.47.$

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Question 873 Marks

If the mean of the following data is 18.75. Find the value of p.

x	10	15	P	25	30
f	5	10	7	8	2

Answer

x	f	fx
10	5	50
15	10	150
P	7	7P
25	8	200
30	2	60
	N = 32	$\sum\text{fx}=7\text{P}+460.$

$\Rightarrow\text{Given mean=18.75}$
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=18.75$
$\Rightarrow\frac{7\text{P}+460}{32}=18.75$
$\Rightarrow7\text{P}+460=600$
$\Rightarrow7\text{P}=140$
$\Rightarrow\text{P}=\frac{140}{7}$
$\Rightarrow\text{P}=20$

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Question 883 Marks

Find the mean from the following frequency distribution of marks at a test in statistics:

Marks (x)	5	10	15	20	25	30	35	40	45	50
No. of students (f)	15	50	80	76	72	45	39	9	8	6

Answer

Let assumed mean = 25

Marks (x)	No. of students (f)	d = x - A (A = 25)	fd
5	15	-20	-300
10	50	-15	-750
15	80	-10	-800
20	76	- 5	-380
25 - A	72	0	0
30	45	5	245
35	39	10	390
40	9	15	135
45	8	20	160
50	6	25	150
Total	400		-1170

$\text{Mean}=\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}=25+\frac{-1170}{400}$

$=25-\frac{117}{40}=25-2.925=22.075$

$\therefore$ Average of marked obtained per student = 22.075

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Question 893 Marks

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days student was absent.

Number of days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
Number of students	11	10	7	4	4	3	1

Answer

No. of days	Class Marks (x)	No. of students (f)	fx
0-6	3	11	33
6-10	8	10	80
10-14	12	7	84
14-20	17	4	68
20-28	24	4	96
28-38	33	3	99
38-40	39	1	39
Total		40	499

Mean $= \frac{\sum\text{f}\ \text{x}}{\sum\ \text{f}}=\frac{499}{40}=12.475$
$\therefore$ Mean number of days a students was absent = 12.475

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Question 903 Marks

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes	$50-52$	$53-55$	$56-58$	$59-61$	$62-64$
Number of boxes	$15$	$110$	$135$	$115$	$25$

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer

We shall apply the assumed mean deviation method
Let assumed mean $(A) = 57$
We shall choose the method of assumed mean deviation:

No. of mangoes	Class marks $(x)$	No. of boxes $(f)$	$d_i = x_i - A A = 57$	$\text{u}_\text{i}=\frac{\text{d}_\text{i}}{\text{h}}$ $\text{h}=3$	$f_i u_i$
$50-52$	$51$	$15$	$-6$	$-2$	$-30$
$53-55$	$54$	$110$	$-3$	$-1$	$-110$
$56-58$	$57$	$135$	$0$	$0$	$0$
$59-61$	$60$	$115$	$3$	$1$	$115$
$61-64$	$63$	$25$	$6$	$2$	$50$
Total		$400$			$25$

$\text{Mean}=\text{A}+\text{h}\times\frac{{\sum\text{f}_\text{i}\text{u}_\text{i}}}{\sum\text{f}_\text{i}}$
$=57+3\times\frac{25}{100}$
$=57+\frac{3}{16}$
$=57+0.1875=57.1875=57.19$

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Question 913 Marks

In the first proof reading of a book containing $300$ pages the following distribution of misprints was obtained:

No. of misprints per page $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
No. of page $(f)$	$154$	$95$	$36$	$9$	$5$	$1$

Find the average number of misprints per page.

Answer

Let the assumed mean $(A) = 2$

No. of misprints per page $(x_i)$	No. of days $(f_i)$	$u_i = x_i - A = x _i - 2$	$f_iu_i$
$0$	$154$	$-2$	$-308$
$1$	$95$	$-1$	$-95$
$2$	$36$	$0$	$0$
$3$	$9$	$1$	$9$
$4$	$5$	$2$	$10$
$5$	$1$	$3$	$3$
	N = 300		$\sum\text{f}_\text{i}\text{u}_\text{i}=-381$

Average number of mis prints per day $=\text{A}+\frac{\text{f}_\text{i}\text{u}_\text{i}}{\text{N}}$
$=2-\frac{381}{300}$
$=2-\frac{381}{300}$
$=\frac{600-381}{300}$
$=\frac{219}{300}$
$=0.73$

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Question 923 Marks

The number of students absent in a class were recorded every day for $120$ days and the:

No of Students absent $(x)$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
No of days $(y)$	$1$	$4$	$10$	$50$	$34$	$15$	$4$	$2$

Answer

Let assumed mean $= 3$

Number of students absent $(x)$	Number of days $(f)$	$d = x - A, A = 3$	$f_id_i$
$0$	$1$	$-3$	$-3$
$1$	$4$	$-2$	$-8$
$2$	$10$	$-1$	$-10$
$3$	$50$	$0$	$0$
$4$	$34$	$1$	$34$
$5$	$15$	$2$	$30$
$6$	$4$	$3$	$12$
$7$	$2$	$4$	$8$
Total	$120$		$63$

$\text{Mean}=\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}}=3+\frac{63}{120}$
$=3+0.525=3.525=3.53\ (\text{approx})$

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Question 933 Marks

Find the mean of each of the following frequency distributions:

Class interval	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$9$	$12$	$15$	$10$	$14$

Answer

Let assumed mean $(A) = 25$

Class interval	Class Marks$(x)$	Frequency$(f)$	$d = x - A$	$f_id_i$
$0-10$	$5$	$9$	$-20$	$-180$
$10-20$	$15$	$12$	$-10$	$-120$
$20-30$	$25 - A$	$15$	$0$	$0$
$30-40$	$35$	$10$	$10$	$100$
$40-50$	$45$	$14$	$20$	$280$
Total		$60$		$80$

$\text{Mean}=\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}}=25+\frac{80}{60}$
$=25+\frac{4}{3}=25+1.333=26.333$

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Question 943 Marks

The following table gives the number of children of $150$ families in a village:

No of children $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
No of familles $(f)$	$10$	$21$	$55$	$42$	$15$	$7$

Find the average number of children per family.

Answer

Let the assured mean $(A) = 2$

No of children $(x_i)$	No of faimlies $(f_i)$	$u_i = x_{i }- A = x_i = 2$	$f_iu_i$
$0$	$10$	$-2$	$-20$
$1$	$21$	$-1$	$-21$
$2$	$55$	$0$	$0$
$3$	$42$	$1$	$42$
$4$	$15$	$2$	$30$
$5$	$7$	$3$	$21$
	$N = 150$		$\sum\text{f}_\text{i}\text{u}_\text{i}=52$

$\therefore$ Average number of children for family $=\text{A}+\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\text{N}}$
$=2+\frac{52}{150}$
$=\frac{300+52}{150}$
$=\frac{352}{150}$
$=2.35(\text{approx})$

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Question 953 Marks

The marks scored by 750 students in an examination are given in the form of a frequency distribution table:

Marks	No. of students
600-640	16
640-680	45
680-720	156
720-760	284
760-800	172
800-840	59
840-880	18

Prepare a cumulative frequency table by less than method and
draw on ogive.

Answer

We first prepare the cumulative frequency table by less than method as given below

Marks	No. of students	Marks less than	Cumulative frequency
600-640	16	640	16
640-680	45	680	61
680-720	156	720	217
720-760	284	760	501
760-800	172	800	673
800-840	59	840	732
840-880	18	880	750

Now, we mark the upper class limits along x-axis and cumulative frequency along y-axis on a suitable gear. Thus, we plot the points (640, 16), (680, 61), (720, 217), (760, 501), (800, 673), (840, 732) and (880, 750)

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Question 963 Marks

Find the mean of each of the following frequency distributions:

Class interval	$10-30$	$30-50$	$50-70$	$70-90$	$90-110$	$110-130$
Frequency	$5$	$8$	$12$	$20$	$3$	$2$

Answer

Let assured mean $(A) = 60$

Class interval	Class Marks$(x)$	Frequency$(f)$	$d = x - A$	$f_id_i$
$10-30$	$20$	$5$	$-40$	$-200$
$30-50$	$40$	$8$	$-20$	$-160$
$50-70$	$60$	$12$	$0$	$0$
$70-90$	$80$	$20$	$20$	$400$
$90-110$	$100$	$3$	$40$	$120$
$110-130$	$120$	$2$	$60$	$120$
Total		$50$		$280$

$\text{Mean}=\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}}=60+\frac{280}{50}$
$=60+\frac{28}{5}=60+5.6=65.6$

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Maths 3 Marks Question