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Mathematics [3 marks sum]

Statistics · ICSE STD 10 (ICSE - English Medium). 23 questions.

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[3 marks sum]

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23 questions · timed · auto-graded

Question 13 Marks
The Mean of n observation $x_1, x_2,..., x_n$ is $\overline{ X }$. $f (a - b)$ is added to each of the observation, show that the mean of the new set of observation is $\overline{ X } + (a - b).$
Answer
We have
$\overline{ X }=\frac{x_1+x_2+\ldots+x_{ n }}{ n } ...(i)$
Let $\bar{X}$ be the mean of $x_1+(a-b), x_2+(a-b), \ldots, x_n+(a-b)$. Then
$\overline{ X }=\frac{\left[x_1+(a-b)\right]+\left[x_2+(a-b)\right]+\ldots+\left[x_2+(a-b)\right]}{ n }$
$=\frac{x_1+x_2+\ldots+x_{ n }+ n (a-b)}{ n }$
$=\frac{x_1+x_2+\ldots+x_{ n }}{ n }+\frac{ n (a-b)}{ n }$
$=\overline{ X }+(a-b) . \quad \ldots[\text { Using (i) }]$
Hence proved.
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Question 23 Marks
Following table present educational level (middle stage) of females in Arunachal pradesh according to 1981 census:
Age groupNumber of females
(to the nearest ten)
10 - 14300
15 - 19980
20 - 24800
25 - 29380
30 - 34290
Draw a histogram to represent the above data.
Answer
Let us convert the given class intervals into continuous class intervals. Then the given frequency distribution takes the form:
Age groupNumber of females
(to the nearest ten)
9·5 - 14·5300
14·5 - 19·5980
19·5 - 24·5800
24·5 - 29·5380
29·5 - 34·5290
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Question 33 Marks
Draw a histogram to represent the following data:
Pocket money in ₹No. of Students
150 - 20010
200 - 2505
250 - 3007
300 - 3504
350 - 4003
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Question 43 Marks
Draw the histogram for the following frequency distribution and hence estimate the mode for the distribution.
ClassFrequency
0 - 52
5 - 107
10 - 1518
15 - 2010
20 - 258
25 - 305
Total24
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Question 53 Marks
Draw a histogram and frequency polygon to represent the following data (on the same scale) which shows the monthly cost of living index of a city in a period of 2 years:
Cost of living IndexNumber of months
440 - 4602
460 - 4804
480 - 5003
500 - 5205
520 - 5403
540 - 5602
560 - 5801
580 - 6004
Total 24
Answer
Histogram and frequency polygon representing the cost of living index of city in a period of 2 years:
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Question 63 Marks
Distribution of height in cm of 100 people is given below:
Class interval (cm)Frequency
145 - 1553
155 - 16535
165 - 17525
175 - 18515
185 - 19520
195 - 2052
Draw a histogram to represent the above data.
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Question 73 Marks
From the following numbers find the median:
$10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9, 15.$
Answer
On arranging in ascending order
$3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81$
Here, $n = 12$ which is even
Therefore, median $=\frac{\left(\frac{ n }{2}\right)^{\text {th }} \operatorname{term}+\left(\frac{ n }{2}+1\right)^{\text {th }} \text { term }}{2}$
$=\frac{\left(\frac{12}{2}\right)^{\text {th }} \text { term }+\left(\frac{12}{2}+1\right)^{\text {th }} \text { term }}{2} $
$ =\frac{6^{\text {th }} \text { term }+7^{\text {th }} \text { term }}{2}$
$=\frac{15+17}{2} $
$ =\frac{32}{2}=16$
Median $= 16.$
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Question 83 Marks
There are $50$ students in a class in which $40$ are boys and rest are girls. The average weight of the class is $44$ kgs and the average weight of the girls is $40$ kgs. Find the average weight of the boys.
Answer
We have
$n =$ No., od students in a class $= 50$
$n_1 =$ No., of boys in a class $= 40$
$n_2 =$ No., of girls in a class $= 10$
$\overline{ X }_1=$ Average weight of boys $= ?$
$\overline{ X }_2=$ Average weight of girls $=40 kgs.$
$\therefore \overline{ X }=\frac{ n _1 \overline{ X }_1+ n _2 \overline{ X }_2}{ n _1+ n _2}$
$\Rightarrow 44=\frac{40 \overline{ X }_1+10 \times 40}{40+10}$
$\Rightarrow 50 \times 44=40 \overline{ X }_1+400$
$\Rightarrow 2200=40 \overline{ X }+400$
$\Rightarrow 40 \overline{ X }_1=1800$
$\Rightarrow \overline{ X }_1=45$
Hence, the average weight of boys is $45$ kgs.
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Question 93 Marks
The average score of boys in an examination of a school is $71$ and of girls is $73.$ The averages score of school in that examination is $71.8.$ Find the ratio of the number of boys between number of girls appeared in the examination.
Answer
Let $\overline{ X }_1$ and $\overline{ X }_2$ be the average scores of boys and girls respectively and $\overline{ X }$ be the average of both boys and girls.
Then
$\overline{ X }_1=71, \overline{ X }_2=73, \overline{ X }=71 \cdot 8$
$\therefore \overline{ X }=\frac{ n _1 \overline{ X }_1+ n _2 \overline{ X }_2}{ n _1+ n _2}$
$\Rightarrow 71 \cdot 8=\frac{ n _1 \times 71+ n _2 \times 73}{ n _1+ n _2}$
$\Rightarrow 71·8 (n_1 + n_2) = 71n_1 + 73n_2$
$\Rightarrow 0·8n_1 = 1·2n_2$
$\Rightarrow 8n_1 = 12n_2$
$\Rightarrow \frac{ n _1}{ n _2}=\frac{12}{8}=\frac{3}{2}$
Hence $n_1 : n_2 = 3 : 2.$
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Question 103 Marks
In X standard, there are three sections A, B and C with $25, 40$ and $35$ students respectively. The average marks of section A is $70\%,$ section B is $65\%$ and of section C is $50\%.$ Find the average marks of the entire X standard.
Answer
Here, $n_1 = 25, n_2 = 40, n_3 = 35,$
$\overline{ X }_1=70, \overline{ X }_2=65$ and $\overline{ X }_3=50$
Let $\overline{ X }$ denote the average marks of the entire $X$ standard.
Then,
$\overline{ X }=\frac{ n _1 \overline{ X }_1+ n _2 \overline{ X }_2+ n _3 \overline{ X }_3}{ n _1+ n _2+ n _3}$
$=\frac{25 \times 70+40 \times 65+35 \times 50}{25+40+35}$
$=\frac{1750+2600+1750}{100}$
$=\frac{6100}{100}=61$
Hence, the average marks of the entire X standard is $61\%.$
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Question 113 Marks
Find the mode for the following series:
2.5, 2.3, 2.2, 2.2, 2.4, 2.7, 2.7, 2.5, 2.3, 2.2, 2.6, 2.2.
Answer
Arranging the data in the form of a frequency table, we have:
ValueTally barsFrequency
2·2IIII4
2·3II2
2·4I1
2·5II2
2·6I1
2·7II2
We see that the value 2·2 has the maximum frequency i.e., 4.
So, 2·2 is the mode for the given series.
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Question 123 Marks
The median of the following observations arranged in ascending order is $24.$ Find $x:$
$11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41.$
Answer
$11, 12, 14, 18, x + 2, x + 4, 30, 32, 35$
$n = 10$ (even), Median $= 24$
$\therefore$ Median $=\frac{\left(\frac{ n }{2}\right)^{\text {th }} \text { term }+\left(\frac{ n }{2}+1\right)^{\text {th }} \text { term }}{2}$
$=\frac{\left(\frac{10}{2}\right)^{\text {th }} \text { term }+\left(\frac{10}{2}+1\right)^{\text {th }} \text { term }}{2} $
$ \text { Median }=\frac{5^{\text {th }} \text { term }+6^{\text {th }} \text { term }}{2}$
$24=\frac{x+2+x+4}{2} $
$ 2 x+6=24 \times 2 $
$ \Rightarrow 2 x=48-6 $
$ \Rightarrow 2 x=42 $
$ x=21 .$
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Question 133 Marks
Find the Median of the following data:
$2,10,9,9,5,2,3,7,11,15.$
Answer
Arranging the data in ascending order, we get
$2, 2, 3, 5, 7, 9, 9, 10, 11, 15$
Here $n = 10$ (Even)
So, Median
$=\frac{\left(\frac{ n }{2}\right)^{\text {th }} \text { term }+\left(\frac{ n }{2}+1\right)^{ th } \text { term }}{2} \\ =\frac{\left(\frac{10}{2}\right)^{\text {th }} \text { term }+\left(\frac{10}{2}+1\right)^{\text {th }} \text { term }}{2}$
$ =\frac{5^{\text {th }} \text { term }+6^{\text {th }} \text { term }}{2} $
$=\frac{7+9}{2}$
$=\frac{16}{2}$
$= 8.$
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Question 143 Marks
Find the Median of the following data:
$12,17, 3,14, 6, 9,8,15,20$
Answer
Arranging the data in ascending order, we get
$3, 6, 8, 9, 12, 14, 15, 17, 20$
Here, $n = 9$ (odd)
Hence,
$\text { Median }=\left(\frac{ n +1}{2}\right)^{\text {th }} \text { item } $
$ =\left(\frac{9+1}{2}\right)^{\text {th }} \text { item }$
$= 5^{th}$ item
$= 12.$
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Question 153 Marks
Find the mean of the following frequency distribution:
Class Interval Frequency
0 - 50 4
50 - 100 8
100 - 150 16
150 - 200 13
200 - 250 6
250 - 300 3
Answer
Class - Interval (x) (f) (fx)
0-50 24 4 100
50-100 75 8 600
100-150 125 16 2,000
150-200 175 13 2,275
200-250 225 6 1,350
250-300 275 3 825
    $\sum f= 5 0$ $\sum f x=7,150$
$\therefore \text { Mean }=\frac{\sum f x}{\sum f}$
$=\frac{7150}{50}$
$=143 .$
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Question 163 Marks
Find the mean of the following distribution:
Class interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
Frequency 10 6 8 12 5
Answer
Class
Interval		 Frequency
(f)		 Mid value
x		 fx
0-10 10 5 50
Oct-20 6 15 90
20-30 8 25 200
30-40 12 35 420
40-50 5 45 225
  $\sum f=41$   $\sum f x=985$
$\therefore \text { Mean }=\frac{\sum f x}{\sum f}=\frac{985}{41}$
$=24.02$
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Question 173 Marks
Find the mean of the following distribution:
x 10 30 50 70 89
f 7 8 10 15 10
Answer
Marks
(x)		 Number of
students (f)		 fx
5 6 30
6 a 6a
7 16 112
8 13 104
9 b 9b
  $\sum f=34+a+b$ $\sum f x=246+6 a+9 b$
It is given that the number of students is $40.$
$\therefore 35 + a + b = 40$
$\rightarrow a + b – 5 = 0 ....(1)$
$\text { Mean }=\frac{\sum f x}{\sum f}$
$\Rightarrow \frac{246+6 a+9 b}{35+a+b}=7.2$
$\Rightarrow 246+6 a+9 b=7.2(35+a+b)$
$\Rightarrow 246+6 a+9 b=252+7.2 a+7.2 b$
$\Rightarrow 0=252-246+7.2 a-6 a+7.2 b-9 b$
$\Rightarrow 6+1.2 a-1.8 b=0$
$\Rightarrow 10+2 a-3 b=0 \quad \ldots (2)$
Solving equations $(1)$ and $(2),$ we have
$5a - 5 = 0$
$\Rightarrow a=1$
From $(1),$ we have $b = 4$
Hence, the values of $a$ and $b$ are $1$ and $4$ respectively.
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Question 183 Marks
Find the mean of the following distribution:
x 4 6 9 10 15
f 5 10 10 7 8
Answer
Calculation of Arithmetic MEan:
$x_i$ $f_i$ $f_i x_i$
10 7 70
30 8 240
50 10 500
70 15 1050
59 10 890
  $\sum f_i=N=50$ $\sum f_i x_i=2750$
$\therefore \text { Mean }=\frac{\sum f_i x_i}{N}$
$=\frac{2750}{50}$
$=55 .$
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Question 193 Marks
Obtain the median for the following frequency distribution:
x :123456789
f :810111620251596
Answer
Calculation of Median
Image
Here, N = 120, so, $\frac{N}{2}=60$.
The cumulative frequency just greater than $\frac{N}{2}$ i.e., 60 is 65 . The value of the variate corresponding to 65 is 5.
Hence, median = 5.
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Question 203 Marks
Calculate the median of the following distribution:
Weight (in nearest kg.) No. of students
$46$ $7$
$48$ $5$
$50$ $8$
$52$ $12$
$53$ $10$
$54$ $2$
$55$ $1$
Answer
The given variates (weights of students) are already in ascending order. We construct the cumulative frequency table as under:
Variable
(weight)		 Frequency
(No.of Students)		 Cumulative
frequency
$46$ $7$ $7$
$48$ $5$ $12$
$50$ $8$ $20$
$52$ $12$ $32$
$53$ $10$ $42$
$54$ $2$ $44$
$55$ $1$ $45$
Here, $n = 45$, which is odd.
$\therefore$ Median $=\frac{ n +1}{2}$ th observation
$=23^{\text {rd }}$ observation $=52$.
$\left(\because\right.$ All observation from $21^{\text {st }}$ to $32^{\text {nd }}$ are equal, each $=52$ ).
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Question 213 Marks
Find the median of the following frequency distribution:
x 10 11 12 13 14 15
f 1 4 7 5 9 3
Answer
x f c.f.
10 1 1
11 4 5
12 7 12
13 5 17
14 9 26
15 3 29
$n = 29$ (odd)
$\therefore$ Median = $\left(\frac{ n +1}{2}\right)^{\text {th }}$ value
$=\left(\frac{29+1}{2}\right)^{\text {th }} \text { value }$
$=15 \text { th value }$
$=13$
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Question 223 Marks
Find the mode of the following frequency distribution:
x101112131415
f147593
Answer
xf c.f.
1011
1145
12712
13517
14926
15329
⇒ Mode = 14 ...(Since 14 has highest frequency)
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Question 233 Marks
The contents of 100 match box were checked to determine the number of match sticks they contained.
Number of match sticksNumber of boxes
356
3610
3718
3825
3921
4012
418
(i) Calculate correct to one decimal place, the mean number of match sticks per box.
(ii) Determine how many matchsticks would have to be added. To the total contents of the 100 boxes to bring the mean up exactly 39 match sticks.
Answer
Number of match sticks
$\left(x_i\right)$		Number of boxes
$\left(f_i\right)$		$f_i x_i$
356210
3610360
3718666
3825950
3921819
4012480
418328
$\sum f_i=100$$\sum f_i x_i=3813$
$\therefore$ Mean $=\frac{\sum f_i x_i}{\sum f_i}=\frac{3813}{100}$
= 38.13 ∼ 38.1.
(ii) Now the number of extra sticks to be added.
= 39 x 100 - 38.13 x 100
= 3900 - 3813 = 87.
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