Math Ncert Exemplar 2019 Solutions for Class 9 Maths Chapter 14 Statistics And Probability are provided here with simple stepbystep explanations. These solutions for Statistics And Probability are extremely popular among Class 9 students for Maths Statistics And Probability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of Class 9 Maths Chapter 14 are provided here for you for free. You will also love the adfree experience on Meritnation’s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.
Page No 131:
Question 1:
Write the correct answer in each of the following :
The class mark of the class 90120 is :
(A) 90
(B) 105
(C) 115
(D) 120
Answer:
Class mark is the mean of the upper and lower limit of class interval.
$=\frac{90+120}{2}\phantom{\rule{0ex}{0ex}}=105$
Hence, the correct answer is option B.
Page No 131:
Question 2:
Write the correct answer in each of the following :
The range of the data :
25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is
(A) 10
(B) 15
(C) 18
(D) 26
Answer:
We know,
Range of data = Maximum value – Minimum value
Now, maximum value = 32, minimum value = 6
Range of data = 32 – 6
Hence, the correct answer is option D.
Page No 131:
Question 3:
Write the correct answer in each of the following :
In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is :
(A) 6
(B) 7
(C) 8
(D) 12
Answer:
Let x and y be the upper and lower class limit.
Now, midvalue of class
$\frac{x+y}{2}=10\left(\mathrm{given}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow x+y=20.....\left(1\right)$
Width of class is 6
i.e. x – y = 6 .....(2)
Adding (1) and (2), we get
2x = 26
⇒ x = 13
Also, y = 7
Here, lower limit of class is 7.
Hence, the correct answer is option B.
Page No 132:
Question 4:
Write the correct answer in each of the following :
The width of each of five continuous classes in a frequency distribution is 5 and the lower classlimit of the lowest class is 10. The upper classlimit of the highest class is:
(A) 15
(B) 25
(C) 35
(D) 40
Answer:
Lower class limit = 10
Width of each class = 5
Width till upper class limit for a frequency distribution having 5 classes = 5 × 5
Hence, the correct answer is option C.
Page No 132:
Question 5:
Write the correct answer in each of the following :
Let m be the midpoint and l be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is :
(A) 2m + l
(B) 2m – l
(C) m – l
(D) m – 2l
Answer:
Let 'a' be lower class limit, l be upper class limit and 'm' be midpoint.
$\therefore m=\frac{a+l}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2m=a+l\phantom{\rule{0ex}{0ex}}\Rightarrow a=2ml$
Hence, the correct answer is option B.
Page No 132:
Question 6:
Write the correct answer in each of the following :
The class marks of a frequency distribution are given as follows :
15, 20, 25, ...
The class corresponding to the class mark 20 is :
(A) 12.5 – 17.5
(B) 17.5 – 22.5
(C) 18.5 – 21.5
(D) 19.5 – 20.5
Answer:
The class marks of frequency distribution are given as 15, 20, 25....
Now, class size = 20 – 15
$\begin{array}{rcl}\mathrm{Upper}\mathrm{limit}& =& \mathrm{Class}\mathrm{mark}+\frac{\mathrm{Class}\mathrm{size}}{2}\\ & =& 20+\frac{5}{2}\\ & =& 22.5\end{array}$
$\begin{array}{rcl}\mathrm{Lower}\mathrm{limit}& =& \mathrm{Class}\mathrm{mark}\frac{\mathrm{Class}\mathrm{size}}{2}\\ & =& 20\frac{5}{2}\\ & =& 17.5\end{array}$
Thus, 17.5 – 22.5 is the corresponding class to the class mark 20.
Hence, the correct answer is option B.
Page No 132:
Question 7:
Write the correct answer in each of the following :
In the class intervals 10  20, 20  30, the number 20 is included in :
(A) 10 – 20
(B) 20 – 30
(C) both the intervals
(D) none of these intervals
Answer:
Here, 20 will be included in interval 20  30.
Hence, the correct answer is option B.
Page No 132:
Question 8:
Write the correct answer in each of the following :
A grouped frequency table with class intervals of equal sizes using 250 – 270 (270 not included in this interval) as one of the class interval is constructed for the following data :
268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236.
The frequency of the class 310330 is:
(A) 4
(B) 5
(C) 6
(D) 7
Answer:
Numbers between 310 and 330 are 310, 320, 319, 310, 318, 316. 330 will not be included in this interval.
Here, frequency is 6.
Hence, the correct answer is option C.
Page No 132:
Question 9:
Write the correct answer in each of the following :
A grouped frequency distribution table with classes of equal sizes using 63  72 (72 included) as one of the class is constructed for the following data :
30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, 40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96, 102, 110, 88, 74, 112, 14, 34, 44.
The number of classes in the distribution will be :
(A) 9
(B) 10
(C) 11
(D) 12
Answer:
Here, the frequency varies from 14 to 112
So, class intervals will be 13  22, 23  32, 33  42, 43  52, 53  62, 63  72, 73  82, 83  92, 93  102, 103  112.
Thus, the number of class intervals = 10.
Hence, the correct answer is option B.
Page No 132:
Question 10:
Write the correct answer in each of the following :
To draw a histogram to represent the following frequency distribution :
Class interval  5 – 10  10 – 15  15 – 25  25 – 45  45 – 75 
Frequency  6  12  10  8  15 
the adjusted frequency for the class 25 – 45 is :
(A) 6
(B) 5
(C) 3
(D) 2
Answer:
The frequency distribution will consist of classes with an interval of 5.
The class 25 – 45 includes four such classes with an interval of 5.
Thus, adjusted frequency will be $\frac{8}{4}=2$.
Hence, the correct answer is option D.
Page No 133:
Question 11:
Write the correct answer in each of the following :
The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is :
(A) 28
(B) 30
(C) 35
(D) 38
Answer:
Let five numbers be a, b, c, d, e.
Then,
$\frac{a+b+c+d+e}{5}=30\phantom{\rule{0ex}{0ex}}\Rightarrow a+b+c+d+e=150.....\left(1\right)$
Let the number excluded be 'a'.
$\Rightarrow \frac{b+c+d+e}{4}=28\phantom{\rule{0ex}{0ex}}\Rightarrow b+c+d+e=112\phantom{\rule{0ex}{0ex}}\Rightarrow a+112=150\left(\mathrm{from}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}\Rightarrow a=38$
Hence, the correct answer is option D.
Page No 133:
Question 12:
Write the correct answer in each of the following :
If the mean of the observations :
x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is
(A) $10\frac{1}{3}$
(B) $10\frac{2}{3}$
(C) $11\frac{1}{3}$
(D) $11\frac{2}{3}$
Answer:
Mean of x, x + 3, x + 5, x + 7, x + 10 is 9.
$\Rightarrow \mathrm{Mean}=\frac{5x+25}{5}=9\phantom{\rule{0ex}{0ex}}\Rightarrow 5x+25=45\phantom{\rule{0ex}{0ex}}\Rightarrow 5x=20\phantom{\rule{0ex}{0ex}}\Rightarrow x=4$
$\begin{array}{rcl}\therefore \mathrm{Mean}\mathrm{of}\mathrm{last}\mathrm{three}\mathrm{terms}& =& \frac{9+11+14}{3}\\ & =& \frac{34}{3}\\ & =& 11\frac{1}{3}\end{array}$
Hence, the correct answer is option C.
Page No 133:
Question 13:
Write the correct answer in each of the following :
If $\overline{x}$ represents the mean of n observations x_{1}, x_{2}, ..., x_{n}, then value of $\sum _{i=1}^{n}\left({x}_{i}\overline{x}\right)$is:
(A) –1
(B) 0
(C) 1
(D) n – 1
Answer:
Given,
$\overline{)x}=\frac{{\displaystyle \sum _{i=1}^{n}}{x}_{i}}{n}\phantom{\rule{0ex}{0ex}}\Rightarrow n\overline{)x}=\sum _{i=1}^{n}{x}_{i}.....\left(1\right)$
$\begin{array}{rcl}\mathrm{Now},\sum _{i=1}^{n}\left({x}_{i}\overline{)x}\right)& =& \sum _{i=1}^{n}{x}_{i}\sum _{i=1}^{n}\overline{)x}\\ & =& n\overline{)x}\overline{)x}\underset{i=1}{\overset{n}{\sum 1}}\left(\mathrm{from}\left(1\right)\right)\\ & =& n\overline{)x}\overline{)x}n\left(\because \sum _{i=1}^{n}\left(1\right)=n\right)\\ & =& 0\end{array}$
Hence, the correct answer is option B.
Page No 133:
Question 14:
Write the correct answer in each of the following :
If each observation of the data is increased by 5, then their mean
(A) remains the same
(B) becomes 5 times the original mean
(C) is decreased by 5
(D) is increased by 5
Answer:
Let x_{1}, x_{2}, ......x_{n} be n observations
${\overline{)x}}_{\mathrm{Old}}=\frac{{x}_{1}+{x}_{2}+......{x}_{n}}{n}.....\left(1\right)$
Now, adding 5 each observations, we get
${\overline{)x}}_{\mathrm{New}}=\frac{\left({x}_{1}+5\right)+\left({x}_{2}+5\right)+......\left({x}_{n}+5\right)}{n}\phantom{\rule{0ex}{0ex}}{\overline{)x}}_{\mathrm{New}}=\frac{\left({x}_{1}+{x}_{2}+......{x}_{n}\right)+5n}{n}\phantom{\rule{0ex}{0ex}}{\overline{)x}}_{\mathrm{New}}=\frac{\left({x}_{1}+{x}_{2}+......{x}_{n}\right)}{n}+5\phantom{\rule{0ex}{0ex}}{\overline{)x}}_{\mathrm{New}}={\overline{)x}}_{\mathrm{Old}}+5\left(\mathrm{from}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}$
Thus, new mean increased by 5
Hence, the correct answer is option D.
Page No 133:
Question 15:
Write the correct answer in each of the following :
Let $\overline{)x}$ be the mean of x_{1}, x_{2}, ... , x_{n }and $\overline{)y}$ the mean of y_{1}, y_{2}, ... , y_{n}. If $\overline{)z}$ is the mean of x_{1}, x_{2}, ... , x_{n}, y_{1}, y_{2}, ... , y_{n}, then $\overline{)z}$ is equal to
(A) $\overline{)x}+\overline{)y}$
(B) $\frac{\overline{)x}+\overline{)y}}{2}$
(C) $\frac{\overline{)x}+\overline{)y}}{n}$
(D) $\frac{\overline{)x}+\overline{)y}}{2n}$
Answer:
Given: $\begin{array}{rcl}\overline{)x}& =& \frac{{\displaystyle \sum _{i=1}^{n}{x}_{i}}}{n}\mathrm{and}\overline{)y}=\frac{{\displaystyle \sum _{i=1}^{n}{y}_{i}}}{n}\end{array}$
$\begin{array}{rcl}\mathrm{Now},\overline{)z}& =& \frac{\left({x}_{1}+{x}_{2}+.....{x}_{n}\right)+\left({y}_{1}+{y}_{2}+.....{y}_{n}\right)}{2n}\\ & =& \frac{n\overline{)x}+n\overline{)y}}{2n}\left(\because n\overline{)x}=\sum _{i=1}^{n}{x}_{i},n\overline{)y}=\sum _{i=1}^{n}{y}_{i}\right)\\ & =& \frac{\overline{)x}+\overline{)y}}{2}\end{array}$
Hence, the correct answer is option B.
Page No 133:
Question 16:
Write the correct answer in each of the following :
If $\overline{)x}$ is the mean of x_{1}, x_{2}, ... , x_{n}, then for a ≠ 0, the mean of ax_{1}, ax_{2}, ..., ax_{n}, $\frac{{x}_{1}}{a},\frac{{x}_{2}}{a},...,\frac{{x}_{n}}{a}$ is
(A) $\left(a+\frac{1}{a}\right)\overline{)x}$
(B) $\left(a+\frac{1}{a}\right)\frac{\overline{)x}}{2}$
(C) $\left(a+\frac{1}{a}\right)\frac{\overline{)x}}{n}$
(D) $\frac{\left(a+\frac{1}{a}\right)\overline{)x}}{2n}$
Answer:
$\mathrm{Given}:\overline{)x}=\frac{{\displaystyle \sum _{i=1}^{n}}{x}_{i}}{n}.....\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{mean}\mathrm{of}a{x}_{1},a{x}_{2},.....a{x}_{n},\frac{{x}_{1}}{a},\frac{{x}_{2}}{a},.....\frac{{x}_{n}}{a}\phantom{\rule{0ex}{0ex}}=\frac{a{x}_{1}+a{x}_{2}+.....a{x}_{n}+{\displaystyle \frac{{x}_{1}}{a}}+{\displaystyle \frac{{x}_{2}}{a}}+....{\displaystyle \frac{{x}_{n}}{a}}}{2n}\phantom{\rule{0ex}{0ex}}=\frac{a\left({x}_{1}+{x}_{2}+.....{x}_{n}\right)+{\displaystyle \frac{1}{a}}\left({x}_{1}+{x}_{2}+.....{x}_{n}\right)}{2n}\phantom{\rule{0ex}{0ex}}=\frac{a\left(n\overline{)x}\right)+{\displaystyle \frac{1}{a}}\left(n\overline{)x}\right)}{2n}\phantom{\rule{0ex}{0ex}}=\left(a+\frac{1}{a}\right)\frac{\overline{)x}}{2}$
Hence, the correct answer is option B.
Page No 133:
Question 17:
Write the correct answer in each of the following :
If $\overline{){x}_{1}},\overline{){x}_{2}},\overline{){x}_{3}},...,\overline{){x}_{n}}$ are the means of n groups with n_{1}, n_{2}, ... , n_{n }number of observations respectively, then the mean $\overline{)x}$ of all the groups taken together is given by :
(A) $\sum _{i=1}^{n}{n}_{i}{\overline{)x}}_{i}$
(B) $\frac{{\displaystyle \sum _{i=1}^{n}{n}_{i}{\overline{)x}}_{i}}}{{n}^{2}}$
(C) $\frac{{\displaystyle \sum _{i=1}^{n}{n}_{i}{\overline{)x}}_{i}}}{{\displaystyle \sum _{i=1}^{n}{n}_{i}}}$
(D) $\frac{{\displaystyle \sum _{i=1}^{n}{n}_{i}{\overline{)x}}_{i}}}{2n}$
Answer:
Given, $\overline{){x}_{1}},\overline{){x}_{2}},\overline{){x}_{3}},...,\overline{){x}_{n}}$ are the means of n groups with n_{1}, n_{2}, ... , n_{n}.
$\therefore \overline{{x}_{1}}=\frac{{\displaystyle \sum _{i=1}^{{n}_{1}}}{x}_{i}}{{n}_{1}},\overline{{x}_{2}}=\frac{{\displaystyle \sum _{i=1}^{{n}_{2}}{x}_{i}}}{{n}_{2}},\overline{{x}_{3}}=\frac{{\displaystyle \sum _{i=1}^{{n}_{3}}{x}_{i}}}{{n}_{3}},...,\overline{{x}_{n}}=\frac{{\displaystyle \sum _{i=1}^{{n}_{n}}{x}_{i}}}{{n}_{n}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sum _{i=1}^{{n}_{1}}{x}_{i}={n}_{1}\overline{{x}_{1}},\sum _{i=1}^{{n}_{2}}{x}_{i}={n}_{2}\overline{{x}_{2}},\sum _{i=1}^{{n}_{3}}{x}_{i}={n}_{3}\overline{{x}_{3}},...,\sum _{i=1}^{{n}_{n}}{x}_{i}={n}_{n}\overline{{x}_{n}}$
$\begin{array}{rcl}\therefore \overline{x}& =& \frac{\sum _{i=1}^{{n}_{1}}{x}_{i}+\sum _{i=1}^{{n}_{2}}{x}_{i}+...+\sum _{i=1}^{{n}_{n}}{x}_{i}}{{n}_{1}+{n}_{2}+...+{n}_{n}}\\ & =& \frac{{n}_{1}\overline{{x}_{1}}+{n}_{2}\overline{{x}_{2}}+...+{n}_{n}\overline{{x}_{n}}}{{n}_{1}+{n}_{2}+...+{n}_{n}}\\ & =& \frac{\sum _{i=1}^{{n}_{1}}{n}_{i}\overline{{x}_{i}}}{\sum _{i=1}^{n}{n}_{i}}\end{array}$
Hence, the correct answer is option C.
Page No 134:
Question 18:
Write the correct answer in each of the following :
The mean of 100 observations is 50. If one of the observations which was 50 is replaced by 150, the resulting mean will be :
(A) 50.5
(B) 51
(C) 51.5
(D) 52
Answer:
Given: Mean of 100 observations = 50
∴ Total of observations = 50 × 100
= 5,000 – 50 + 150
= 5,100
$\begin{array}{rcl}\therefore \mathrm{Now},\mathrm{mean}\mathrm{of}\mathrm{observation}& =& \frac{5,100}{100}\\ & =& 51\end{array}$
Hence, the correct answer is option B.
Page No 134:
Question 19:
Write the correct answer in each of the following :
There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be –3.5. The mean of the given numbers is :
(A) 46.5
(B) 49.5
(C) 53.5
(D) 56.5
Answer:
Let x be the mean of 50 numbers.
∴ Sum of 50 numbers = 50x
Since, each number is subtracted from 53
$\Rightarrow \frac{53\times 5050x}{50}=3.5\phantom{\rule{0ex}{0ex}}\Rightarrow 2,65050x=175\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{2,825}{50}=56.5$
Hence, the correct answer is option D.
Page No 134:
Question 20:
Write the correct answer in each of the following :
The mean of 25 observations is 36. Out of these observations if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13^{th }observation is :
(A) 23
(B) 36
(C) 38
(D) 40
Answer:
Mean of 25 observation = 36
Total of 25 observation = 36 × 25
Mean of 13 observation = 32
Sum of 13 observation = 32 × 13
Mean of 13 observation = 40
Sum of 13 observation = 40 × 13
13^{th }observation = (416 + 520) – 900
Hence, the correct answer is option B.
Page No 134:
Question 21:
Write the correct answer in each of the following :
The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is
(A) 45
(B) 49.5
(C) 54
(D) 56
Answer:
Observation in ascending order: 22, 34, 39, 45, 54, 54, 56, 68, 78 and 84
Here total number of observations = 10
Since, n is even,
$\begin{array}{rcl}\mathrm{Median}& =& \frac{{\left({\displaystyle \frac{n}{2}}\right)}^{\mathrm{th}}\mathrm{observation}+\left({\displaystyle \frac{n}{2}}+1\right)\mathrm{observation}}{2}\\ & =& \frac{{\left({\displaystyle \frac{10}{2}}\right)}^{\mathrm{th}}\mathrm{observation}+\left({\displaystyle \frac{10}{2}}+1\right)\mathrm{observation}}{2}\\ & =& \frac{{5}^{\mathrm{th}}\mathrm{observation}+{6}^{\mathrm{th}}\mathrm{observation}}{2}\\ & =& \frac{54+54}{2}\\ & =& 54\end{array}$
Hence, the correct answer is option C.
Page No 134:
Question 22:
Write the correct answer in each of the following :
For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are respectively :
(A) upper limits of the classes
(B) lower limits of the classes
(C) class marks of the classes
(D) upper limits of proceeding classes
Answer:
For drawing a frequency polygon of a continuous frequency distribution, we draw class marks of the class on the abscissa and frequency of class on ordinates.
Also, class marks are mean of lower and upper limit of class intervals.
Hence, the correct answer is option C.
Page No 134:
Question 23:
Write the correct answer in each of the following :
Median of the following numbers :
4, 4, 5, 7, 6, 7, 7, 12, 3 is
(A) 4
(B) 5
(C) 6
(D) 7
Answer:
Arrange in ascending order, we get 3, 4, 4, 5, 6, 7, 7, 7, 12.
Here n = 9 (odd)
$\begin{array}{rcl}\mathrm{Median}& =& {\left(\frac{n+1}{2}\right)}^{\mathrm{th}}\mathrm{observation}\\ & =& {\left(\frac{9+1}{2}\right)}^{\mathrm{th}}\mathrm{observation}\\ & =& {5}^{\mathrm{th}}\mathrm{observation}\\ & =& 6\end{array}$
Hence, the correct answer is option C.
Page No 134:
Question 24:
Write the correct answer in each of the following :
Mode of the data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15 is
(A) 14
(B) 15
(C) 16
(D) 17
Answer:
Arrange in ascending order 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20.
Here, 15 occurs most frequently i.e. 5 times
Thus, mode of data is 15.
Hence, the correct answer is option B.
Page No 134:
Question 25:
Write the correct answer in each of the following :
In a sample study of 642 people, it was found that 514 people have a high school certificate. If a person is selected at random, the probability that the person has a high school certificate is :
(A) 0.5
(B) 0.6
(C) 0.7
(D) 0.8
Answer:
Total number of persons = 642
Number of persons with high school certificate, E = 514
$\begin{array}{rcl}P\left(E\right)& =& \frac{514}{642}\\ & =& 0.80\end{array}$
Hence, the correct answer is option D.
Page No 135:
Question 26:
Write the correct answer in each of the following :
In a survey of 364 children aged 1936 months, it was found that 91 liked to eat potato chips. If a child is selected at random, the probability that he/she does not like to eat potato chips is :
(A) 0.25
(B) 0.50
(C) 0.75
(D) 0.80
Answer:
Here, total cases = 364
Favourable outcomes = 273 (∴ 364 – 91)
$\begin{array}{rcl}\therefore \mathrm{Probability}\mathrm{of}\mathrm{event}& =& \frac{273}{364}\\ & =& 0.75\end{array}$
Hence, the correct answer is option C.
Page No 135:
Question 27:
Write the correct answer in each of the following :
In a medical examination of students of a class, the following blood groups are recorded:
Blood group  A  AB  B  O 
Number of students  10  13  12  5 
A student is selected at random from the class. The probability that he/she has blood group B, is :
(A) $\frac{1}{4}$
(B) $\frac{13}{40}$
(C) $\frac{3}{10}$
(D) $\frac{1}{8}$
Answer:
Total number of students = 10 + 13 + 12 + 5 = 40
Number of students with blood group (B) = 12
$\begin{array}{rcl}\therefore \mathrm{Required}\mathrm{Probability}& =& \frac{12}{40}\\ & =& \frac{3}{10}\end{array}$
Hence, the correct answer is option C.
Page No 135:
Question 28:
Write the correct answer in each of the following :
Two coins are tossed 1000 times and the outcomes are recorded as below :
Number of heads  2  1  0 
Frequency  200  550  250 
Based on this information, the probability for at most one head is
(A) $\frac{1}{5}$
(B) $\frac{1}{4}$
(C) $\frac{4}{5}$
(D) $\frac{3}{4}$
Answer:
Probability of event (at most one Head) = P(0 heads) + P(1 heads)
$=\frac{250}{1,000}+\frac{550}{1,000}\phantom{\rule{0ex}{0ex}}=\frac{800}{1,000}\phantom{\rule{0ex}{0ex}}=0.8\phantom{\rule{0ex}{0ex}}=\frac{4}{5}$
Hence, the correct answer is option C.
Page No 135:
Question 29:
Write the correct answer in each of the following :
80 bulbs are selected at random from a lot and their life time (in hrs) is recorded in the form of a frequency table given below :
Life time (in hours)  300  500  700  900  1100 
Frequency  10  12  23  25  10 
One bulb is selected at random from the lot. The probability that its life is 1150 hours, is
(A) $\frac{1}{80}$
(B) $\frac{7}{16}$
(C) 0
(D) 1
Answer:
Total bulbs = 80
Number of bulbs whose lifetime is 1,150 hours, E = 0
$\begin{array}{rcl}\therefore P\left(E\right)& =& \frac{0}{80}\\ & =& 0\end{array}$
Hence, the correct answer is option C.
Page No 136:
Question 30:
Write the correct answer in each of the following :
80 bulbs are selected at random from a lot and their life time (in hrs) is recorded in the form of a frequency table given below :
Life time (in hours)  300  500  700  900  1100 
Frequency  10  12  23  25  10 
The probability that bulbs selected randomly from the lot has life less than 900 hours is :
(A) $\frac{11}{40}$
(B) $\frac{5}{16}$
(C) $\frac{7}{16}$
(D) $\frac{9}{16}$
Answer:
Total bulbs = 80
Bulbs have lifetime less 900 hours, E = 45
$\begin{array}{rcl}P\left(E\right)& =& \frac{45}{80}\\ & =& \frac{9}{16}\\ & & \end{array}$
Hence, the correct answer is option D.
Page No 136:
Question 1:
The frequency distribution :
Marks  0 – 20  20 – 40  40 – 60  60 – 100 
Number of Students  10  15  20  25 
has been represented graphically as follows :
Do you think this representation is correct? Why?
Answer:
No, the representation is not correct, because, the classes 0  20, 20  40, 40  60, 60  100 are not of uniform width but of varying width.
Page No 137:
Question 2:
In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded :
46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98, 44
Which ‘average’ will be a good representative of the above data and why?
Answer:
Median will be good representative of data,
• Each value occurs once.
• The data is influenced by extreme values.
Page No 137:
Question 3:
A child says that the median of 3, 14, 18, 20, 5 is 18. What doesn’t the child understand about finding the median?
Answer:
Since child says median of 3, 14, 18, 20, 5 is 18, it is clear that the child doesn't understand the fact that the given should be arranged in ascending or descending order.
For finding the median, arrange the data in ascending order 3, 5, 14, 18, 20.
Hence, the median is 14.
Page No 137:
Question 4:
A football player scored the following number of goals in the 10 matches :
1, 3, 2, 5, 8, 6, 1, 4, 7, 9
Since the number of matches is 10 (an even number), therefore, the median
$=\frac{{5}^{\mathrm{th}}\mathrm{observation}+{6}^{\mathrm{th}}\mathrm{observation}}{2}\phantom{\rule{0ex}{0ex}}=\frac{8+6}{2}=7$
Is it the correct answer and why?
Answer:
Since data is not arranged in ascending or descending order, hence it is not correct.
Firstly, arranging data in ascending order, er get 1, 1, 2, 3, 4, 5, 6, 7, 8, 9
Number of observation (n) = 10.
$\begin{array}{rcl}\therefore \mathrm{Median}& =& \frac{{\left({\displaystyle \frac{n}{2}}\right)}^{\mathrm{th}}\mathrm{observation}+{\left({\displaystyle \frac{\mathrm{n}}{2}}+1\right)}^{\mathrm{th}}\mathrm{observation}}{2}\\ & =& \frac{\left({{\displaystyle \frac{10}{2}}}^{\mathrm{th}}\right)\mathrm{observation}+{\left({\displaystyle \frac{10}{2}}+1\right)}^{\mathrm{th}}\mathrm{observation}}{2}\\ & =& \frac{{5}^{\mathrm{th}}\mathrm{observation}+{6}^{\mathrm{th}}\mathrm{observation}}{2}\\ & =& \frac{4+5}{2}\\ & =& 4.5\end{array}$
Page No 137:
Question 5:
Is it correct to say that in a histogram, the area of each rectangle is proportional to the class size of the corresponding class interval? If not, correct the statement.
Answer:
Not correct, because in a histogram, area of each rectangle is proportional to corresponding frequency of its class.
Page No 137:
Question 6:
The class marks of a continuous distribution are :
1.04, 1.14, 1.24, 1.34, 1.44, 1.54 and 1.64
Is it correct to say that the last interval will be 1.55 – 1.73? Justify your answer.
Answer:
Not correct, because the difference between the two consecutive class marks should be equal to class size.
But, here difference between two consecutive marks is 0.1 and class size of 1.55 – 1.73 is 0.18, which is not equal.
Page No 138:
Question 7:
30 children were asked about the number of hours they watched TV programmes last week. The results are recorded as under :
Number of hours  0 5  5  10  10  15  15  20 
Frequency  8  16  4  2 
Can we say that the number of children who watched TV for 10 or more hours a week is 22? Justify your answer.
Answer:
No,
The number of children who watched TV for 10 or more hour in a week are 2 + 4 = 6 hours.
Page No 138:
Question 8:
Can the experimental probability of an event be a negative number? If not, why?
Answer:
No, the experimental probability of an event cannot be a negative number.
This is because,
Number of outcomes favourable to an even (E) ≤ Number of possible outcome.
i.e. 0 ≤ P(E) ≤ 1
Page No 138:
Question 9:
Can the experimental probability of an event be greater than 1? Justify your answer.
Answer:
The experimental probability of an event cannot be greater than 1, since the number of trials in which the event can happen cannot be greater than total number of trials.
Page No 138:
Question 10:
As the number of tosses of a coin increases, the ratio of the number of heads to the total number of tosses will be $\frac{1}{2}$. Is it correct? If not, write the correct one.
Answer:
No, as the number of tosses of a coin increases, the ratio of number of heads to total number of heads will be nearer to $\frac{1}{2}$ not exactly $\frac{1}{2}$.
Page No 140:
Question 1:
The blood groups of 30 students are recorded as follows:
A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A, AB, B, A, A, O, A, AB, B, A, O, B, A, B, A
Prepare a frequency distribution table for the data.
Answer:
It can be observed that 12 students have their blood group as A, 8 as B, 4 as AB and 6 as O.
Therefore, the blood group of 30 students can be represented as
Blood group  Number of students 
A  12 
B  8 
AB  4 
O  6 
Total  30 
Page No 140:
Question 2:
The value of π upto 35 decimal places is given below:
3. 14159265358979323846264338327950288
Make a frequency distribution of the digits 0 to 9 after the decimal point.
Answer:
Frequency distribution table for digits after decimal point of π as follows
Digits  Frequency 
0  1 
1  2 
2  5 
3  6 
4  3 
5  4 
6  3 
7  2 
8  5 
9  4 
Total  35 
Page No 140:
Question 3:
The scores (out of 100) obtained by 33 students in a mathematics test are as follows:
69, 48, 84, 58, 48, 73, 83, 48, 66, 58, 84
66, 64, 71, 64, 66, 69, 66, 83, 66, 69, 71
81, 71, 73, 69, 66, 66, 64, 58, 64, 69, 69
Represent this data in the form of a frequency distribution.
Answer:
The given data in a frequency distribution table can be represented as
Scores  Number of students 
48  3 
58  3 
64  4 
66  7 
69  6 
71  3 
73  2 
81  1 
83  2 
84  2 
Page No 140:
Question 4:
Prepare a continuous grouped frequency distribution from the following data:
Mid  point  Frequency 
5  4 
15  8 
25  13 
35  12 
45  6 
Also find the size of class intervals.
Answer:
We know, if M is the midpoint for a class and h is the class size
Then, lower limit of class interval = $M\frac{h}{2}$.
Upper limit of class interval = $M+\frac{h}{2}$.
Now, class size 15 – 5 = 10.
∴ Class interval formed from midpoint 5
$=\left(5\frac{10}{2}\right)\left(5+\frac{10}{2}\right)$
i.e., 0 – 10
Thus, continuous classes formed
Class Interval  0 – 10  10 – 20  20 – 30  30 – 40  40 – 50 
Frequency  4  8  13  12  6 
Page No 140:
Question 5:
Convert the given frequency distribution into a continuous grouped frequency distribution:
Class interval  Frequency 
150  153  7 
154  157  7 
158  161  15 
162  165  10 
166  169  5 
170  173

6 
In which intervals would 153.5 and 157.5 be included?
Answer:
Here, converting inclusive class interval into exclusive class interval by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit, we get
Class Interval  Frequency 
149.5  153.5  7 
153.5  157.5  7 
157.5  161.5  15 
161.5  165.5  10 
165.5  169.5  5 
169.5  173.5  6 
Thus, 153.5 and 157.5 would be included in the class intervals 153.5  157.5 and 157.5  161.5 respectively.
Page No 141:
Question 6:
The expenditure of a family on different heads in a month is given below:
Head  Food  Education  Clothing  House Rent  Others  Savings 
Expenditure (in Rs)  4000  2500  1000  3500  2500  1500 
Draw a bar graph to represent the data above.
Answer:
The bar graph for given data is shown
Page No 141:
Question 7:
Expenditure on Education of a country during a five year period (2002  2006), in crores of rupees, is given below:
Elementary education  240 
Secondary Education  120 
University Education  190 
Teacher’s Training  20 
Social Education  10 
Other Educational Programmes  115 
Cultural programmes  25 
Technical Education  125 
Represent the information above by a bar graph.
Answer:
The bar graph for given data is shown below
Page No 141:
Question 8:
The following table gives the frequencies of most commonly used letters a, e, i, o, r, t, u from a page of a book :
Letters  a  e  i  o  r  t  u 
Frequency  75  125  80  70  80  95  75 
Represent the information above by a bar graph.
Answer:
Page No 142:
Question 9:
If the mean of the following data is 20.2, find the value of p:
x  10  15  20  25  30 
f  6  8  p  10  6 
Answer:
Variable (x_{i})  Frequency (f_{i})  x_{i}f_{i} 
10 15 20 25 30 
6 8 p 10 6 
60 120 20p 250 180 
Σf_{i }= 30 + p  Σx_{i}f_{i }= 610 + 20p 
Mean = 20.2 (given)
$\therefore \mathrm{Mean}=\frac{\mathrm{\Sigma}{x}_{i}{f}_{i}}{\mathrm{\Sigma}{f}_{i}}\phantom{\rule{0ex}{0ex}}\Rightarrow 20.2=\frac{610+20p}{30+p}$
⇒ 610 + 20p = 606 + 20.2p
⇒ 0.2p = 4
⇒ p = 20
Hence, the value of p is 20.
Page No 142:
Question 10:
Obtain the mean of the following distribution:
Frequency  Variable 
4  4 
8  6 
14  8 
11  10 
3  12 
Answer:
x_{i}  f_{i}  x_{i}f_{i} 
4 6 8 10 12 
4 8 14 11 3 
16 48 112 110 36 
Σf_{i }= 40  Σx_{i}f_{i }= 322 
$\begin{array}{rcl}\mathrm{Now},\mathrm{Mean}& =& \frac{\Sigma {x}_{i}{f}_{i}}{\Sigma {f}_{i}}\\ & =& \frac{322}{40}\\ & =& 8.05\end{array}$
Hence, value of mean is 8.05.
Page No 142:
Question 11:
A class consists of 50 students out of which 30 are girls. The mean of marks scored by girls in a test is 73 (out of 100) and that of boys is 71. Determine the mean score of the whole class.
Answer:
Number of girls = 30
Number of boys = 20 (∵ Total students = 50)
Marks obtained by 30 girls = 30 × 73
Hence, the mean score of the whole class is 72.2.
Page No 142:
Question 12:
Mean of 50 observations was found to be 80.4. But later on, it was discovered that 96 was misread as 69 at one place. Find the correct mean.
Answer:
$\frac{{x}_{1}+{x}_{2}+{x}_{3}+.....{x}_{50}}{50}=80.4\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}+{x}_{2}+.......{x}_{50}=4020.....\left(1\right)$
Let x_{5} = Incorrect value = 69
$\begin{array}{rcl}{x}_{5}^{\text{'}}& =& \mathrm{Correct}\mathrm{value}=96\\ & =& {x}_{5}+27\end{array}$
Adding 27 both sides is equation (1)
${x}_{1}+.....{5}_{x}+27+.....{x}_{50}=4020+27\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}+{x}_{2}+.....{x}_{5}^{\text{'}}+.....{x}_{50}=4047$
$\begin{array}{rcl}\therefore \mathrm{Actual}\mathrm{mean}& =& \frac{{x}_{1}+{x}_{2}+.....{x}_{5}^{\text{'}}+......{x}_{50}}{50}\\ & =& \frac{4097}{50}\\ & =& 80.94\end{array}$
Page No 142:
Question 13:
Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in an ascending order. The median of the data is 24. Find the value of x.
Answer:
Arrange the data is ascending order, 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43.
Total observation, n = 10
$\begin{array}{rcl}\therefore \mathrm{Median}& =& \frac{{\left({\displaystyle \frac{n}{2}}\right)}^{\mathrm{th}}\mathrm{observation}+{\left(\frac{n}{2}+1\right)}^{\mathrm{th}}\mathrm{observation}}{2}\\ & =& \frac{{5}^{\mathrm{th}}\mathrm{observation}+{6}^{\mathrm{th}}\mathrm{observation}}{2}\\ & =& \frac{\left(x+1\right)+\left(2x13\right)}{2}\\ & =& \frac{3x12}{2}\end{array}$
But median = 24 (given)
$\Rightarrow \frac{3x12}{2}=24\phantom{\rule{0ex}{0ex}}\Rightarrow 3x12=48\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=60\phantom{\rule{0ex}{0ex}}\Rightarrow x=20$
Hence, the value of x is 20.
Page No 142:
Question 14:
The points scored by a basket ball team in a series of matches are as follows:
17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, 28
Find the median and mode for the data.
Answer:
Arrange given points in ascending order:
2, 5, 7, 7, 8, 10, 10, 10, 14, 17, 18, 24, 25, 27, 28, 48.
Number of observation (n) = 16
$\begin{array}{rcl}\mathrm{Median}& =& {\left(\frac{n}{2}\right)}^{\mathrm{th}}\mathrm{observation}+{\left(\frac{n}{2}+1\right)}^{\mathrm{th}}\mathrm{observation}\\ & =& \frac{{\left({\displaystyle \frac{16}{2}}\right)}^{\mathrm{th}}\mathrm{observation}+{\left({\displaystyle \frac{16}{2}}+1\right)}^{\mathrm{th}}\mathrm{observation}}{2}\\ & =& \frac{{8}^{\mathrm{th}}\mathrm{observation}+{9}^{\mathrm{th}}\mathrm{observation}}{2}\\ & =& \frac{10+14}{2}\\ & =& 12\end{array}$
Now, mode is the observation which is repeated maximum number of times.
Here, 10 repeated 3 times.
So, Mode = 10
Hence, Median = 12, Mode = 10.
Page No 142:
Question 15:
In the given figure, there is a histogram depicting daily wages of workers in a factory. Construct the frequency distribution table.
Answer:
The frequency distribution table for daily wages and number of workers is
Class Interval  Frequency 
150  200 200  250 250  300 300  350 350  400 
50 30 35 20 10 
Total workers  145 
Hence, the total number of workers is 145.
Page No 143:
Question 16:
A company selected 4000 households at random and surveyed them to find out a
relationship between income level and the number of television sets in a home. The
information so obtained is listed in the following table:
Monthly income
(in Rs)

Number of Televisions/household  
0  1  2  Above 2  
< 10000 10000 – 14999 15000 – 19999 20000 – 24999 25000 and above 
20 10 0 0 0 
80 240 380 520 1100 
10 60 120 370 760 
0 0 30 80 220 
Find the probability:
(i) of a household earning Rs 10000 – Rs 14999 per year and having exactly one television.
(ii) of a household earning Rs 25000 and more per year and owning 2 televisions.
(iii) of a household not having any television.
Answer:
Given: Total events = 4,000
(i) A household earn Rs. 10,000 – Rs. 14,999 per year and have exactly are television.
$\begin{array}{rcl}\mathrm{Required}\mathrm{Probability}& =& \frac{240}{4,000}\\ & =& \frac{3}{50}\end{array}$
(ii) A household earn Rs. 25,000 and more and owing two televisions.
$\begin{array}{rcl}\mathrm{Required}\mathrm{Probability}& =& \frac{760}{4,000}\\ & =& 0.19\end{array}$
Page No 143:
Question 17:
Two dice are thrown simultaneously 500 times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:
Sum  Frequency 
2  14 
3  30 
4  42 
5  55 
6  72 
7  75 
8  70 
9  53 
10  46 
11  28 
12  15 
If the dice are thrown once more, what is the probability of getting a sum
(i) 3?
(ii) more than 10?
(iii) less than or equal to 5?
(iv) between 8 and 12?
Answer:
(i)
$\begin{array}{rcl}P\left(\mathrm{getting}\mathrm{a}\mathrm{sum}3\right)& =& \frac{30}{500}\\ & =& \frac{3}{50}\\ & =& 0.06\end{array}$
$\begin{array}{rcl}P\left(\mathrm{getting}\mathrm{a}\mathrm{sum}\mathrm{more}\mathrm{than}10\right)& =& P\left(\mathrm{getting}\mathrm{a}\mathrm{sum}\mathrm{of}11\right)+P\left(\mathrm{getting}\mathrm{a}\mathrm{sum}12\right)\\ & =& \frac{28}{500}+\frac{15}{500}\\ & =& \frac{43}{500}\\ & =& 0.869\end{array}$
(iii)
$\begin{array}{rcl}P\left(\mathrm{getting}\mathrm{a}\mathrm{sum}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}5\right)& =& P\left(\mathrm{getting}\mathrm{a}\mathrm{sum}\mathrm{of}5\right)+P\left(\mathrm{getting}\mathrm{a}\mathrm{sum}\mathrm{of}4\right)+P\left(\mathrm{getting}\mathrm{a}\mathrm{sum}\mathrm{of}3\right)+P\left(\mathrm{getting}\mathrm{sum}4\right)\\ & =& \frac{55}{500}+\frac{42}{500}+\frac{30}{500}+\frac{14}{500}\\ & =& \frac{141}{500}\\ & =& 0.282\end{array}$
(iv)
$\begin{array}{rcl}P\left(\mathrm{getting}\mathrm{sum}\mathrm{between}8\mathrm{and}12\right)& =& P\left(\mathrm{getting}\mathrm{sum}\mathrm{of}9\right)+P\left(\mathrm{getting}\mathrm{sum}\mathrm{of}10\right)+P\left(\mathrm{getting}\mathrm{a}\mathrm{sum}\mathrm{of}11\right)\\ & =& \frac{53}{500}+\frac{46}{500}+\frac{28}{500}\\ & =& \frac{127}{500}\\ & =& 0.254\end{array}$
Page No 144:
Question 18:
Bulbs are packed in cartons each containing 40 bulbs. Seven hundred cartons were examined for defective bulbs and the results are given in the following table:
Number of defective bulbs  0  1  2  3  4  5  6  more than 6 
Frequency  400  180  48  41  18  8  3  2 
One carton was selected at random. What is the probability that it has
(i) no defective bulb?
(ii) defective bulbs from 2 to 6?
(iii) defective bulbs less than 4?
Answer:
(i) P(no defective bulbs)
$=\frac{400}{700}\phantom{\rule{0ex}{0ex}}=\frac{4}{7}$
(ii) Number of outcomes = 48 + 41 + 18 + 8 + 3 = 118
P(defective bulb from 2 to 6)
$=\frac{118}{700}\phantom{\rule{0ex}{0ex}}=\frac{59}{350}$
(iii) Number of outcomes = 400 + 180 + 48 + 41 = 669
P(defective bulb less than 4)$=\frac{669}{700}$
Page No 144:
Question 19:
Over the past 200 working days, the number of defective parts produced by a machine is given in the following table:
Number of defective parts  0  1  2  3  4  5  6  7  8  9  10  11  12  13 
Days  50  32  22  18  12  12  10  10  10  8  6  6  2  2 
Determine the probability that tomorrow’s output will have
(i) no defective part
(ii) atleast one defective part
(iii) not more than 5 defective parts
(iv) more than 13 defective parts
Answer:
Number of possible outcomes = 200
(i)
$\begin{array}{rcl}P\left(\mathrm{No}\mathrm{defect}\right)& =& \frac{50}{200}\\ & =& 0.25\end{array}$
(ii) The probability of atleast 1 defect = 1 – P(no defect)
= 1 – 0.25
= 0.75
(iii) Probability that there will be maximum 5 defects = P(0 defect) + P(1 defect) + P(2 defect) + P(3 defect) + P(4 defect) + P(5 defect)
(iv) P(more than 13 defective parts) = $\frac{0}{200}=0$
Page No 144:
Question 20:
A recent survey found that the ages of workers in a factory is distributed as follows:
Age (in years)  20 – 29  30 – 39  40 – 49  50 – 59  60 and above 
Number of workers  38  27  86  46  3 
If a person is selected at random, find the probability that the person is:
(i) 40 years or more
(ii) under 40 years
(iii) having age from 30 to 39 years
(iv) under 60 but over 39 years
Answer:
(i) P(person is 40 year or more) = P(person having age 40 to 49 years) + P(person having age 50 to 59 years) + P(person having age 60 and above)
(ii) P(person is under 40 years) = P(person having age 20 to 29 years) + P(person having age 30 to 39 years)
(iii) P(age form 30 to 39 years) = $\frac{27}{200}=0.135$
(iv) P(person having age under 60 but over 39 years) = P(person having age 40 to 19 years) + P(person having age 50 to 59 years)
Page No 147:
Question 1:
The following are the marks (out of 100) of 60 students in mathematics.
16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28,72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63, 25, 36, 54, 44, 47, 27, 72, 17, 4, 30.
Construct a grouped frequency distribution table with width 10 of each class starting from 0  9.
Answer:
Arranging the given data in ascending order.
4, 5, 7, 7,13,15,16,16,17,17,19, 24, 25, 26, 27, 28, 30, 31,31,34, 34, 35, 35, 36, 36, 36, 42, 44, 45, 47, 48, 51, 52, 52, 54, 55, 55, 56, 56, 61,62, 62, 63, 68, 70, 72, 72, 72, 74, 75, 75, 78, 80, 81,85, 86, 86, 92, 95, 97.
We arrange the given data into groups like 0  9, 10  19, 20  29, … The class width in each case is 10.
Class Interval  Frequency 
0  9  4 
10  19  7 
20  29  5 
30  39  10 
40  49  5 
50  59  8 
60  69  5 
70  79  8 
80  89  5 
90  99  3 
Total frequency = 60
Page No 147:
Question 2:
The following are the marks (out of 100) of 60 students in mathematics.
16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28,72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63, 25, 36, 54, 44, 47, 27, 72, 17, 4, 30.
Construct a grouped frequency distribution table with width 10 of each class, in such a way that one of the classes is 10 – 20 (20 not included).
Answer:
The frequency distribution of given data is given below.
Class Interval  Tally Marks  Frequency 
0 – 10  $\left\right\left\right$  4 
10 – 20  $\overline{)\left\right\left\right}\left\right$  7 
20 – 30  $\overline{)\left\right\left\right}$  5 
30 – 40  $\overline{)\left\right\left\right}\overline{)\left\right\left\right}$  10 
40 – 50  $\overline{)\left\right\left\right}$  5 
50 – 60  $\overline{)\left\right\left\right}\left\right$  8 
60 – 70  $\overline{)\left\right\left\right}$  5 
70 – 80  $\overline{)\left\right\left\right}$ $\left\right$  8 
80 – 90  $\overline{)\left\right\left\right}$  5 
90 – 100  $\left\right$  3 
Page No 147:
Question 3:
Draw a histogram of the following distribution :
Heights (in cm)  Number of students 
150 – 153  7 
153 – 156  8 
156 – 159  14 
159 – 162  10 
162 – 165  6 
165 – 168  5 
Answer:
A histogram of the given distribution is given below.
Page No 147:
Question 4:
Draw a histogram to represent the following grouped frequency distribution :
Ages (in years)  Number of teachers 
20  24  10 
25  29  28 
30  34  32 
35  39  48 
40  44  50 
45  49  35 
50  54  12 
Answer:
The given frequency distribution is not continuous. So, first we convert it into continuous form.
Now, consider the class 20  24, 25  29.
Lower limit of 25  29 is 25.
Upper limit of 20  24 is 24.
Thus, the half of the difference is $\frac{2524}{2}=0.5$
So, we subtract 0.5 from each lower limit and add 0.5 to each upper limit.
The table for continuous grouped frequency distribution is given below:
Ages (in years)  Number of teachers 
19.5  24.5  10 
24.5  29.5  28 
29.5  34.5  32 
34.5  39.5  48 
39.5  44.5  50 
44.5  49.5  35 
49.5  54.5  12 
Now, the histogram of the given frequency distribution is:
Page No 148:
Question 5:
The lengths of 62 leaves of a plant are measured in millimetres and the data is represented in the following table :
Length (in mm)  Number of leaves 
118 – 126  8 
127 – 135  10 
136 – 144  12 
145 – 153  17 
154 – 162  7 
163 – 171  5 
172 – 180  3 
Draw a histogram to represent the data above.
Answer:
Histogram of the given distribution is given below
Page No 148:
Question 6:
The marks obtained (out of 100) by a class of 80 students are given below :
Marks  Number of students 
10 – 20  6 
20 – 30  17 
30 – 50  15 
50 – 70  16 
70 – 100  26 
Construct a histogram to represent the data above.
Answer:
Here, $\mathrm{Adjusted}\mathrm{frequency}\mathrm{of}\mathrm{class}=\frac{\mathrm{Minimum}\mathrm{class}\mathrm{size}}{\left(\mathrm{Class}\mathrm{size}\times \mathrm{Frequency}\right)}$
Marks  Number of Students  Adjusted Frequency 
10 – 20  6  $\frac{10}{10}\times 6=6$ 
20 – 30  17  $\frac{10}{10}\times 17=17$ 
30 – 50  15  $\frac{10}{20}\times 15=7.5$ 
50 – 70  16  $\frac{10}{20}\times 16=8$ 
70 – 100  26  $\frac{10}{30}\times 26=8.67$ 
Page No 148:
Question 7:
Following table shows a frequency distribution for the speed of cars passing through at a particular spot on a high way :
Class interval (km/h)  Frequency 
30  40  3 
40  50  6 
50  60  25 
60  70  65 
70  80  50 
80  90  28 
90  100  14 
Draw a histogram and frequency polygon representing the data above.
Answer:
Let us draw a histogram for this data and mark the midpoints of the top of the rectangles as B, C, D, E, F, G and H respectively.
Here, the first class is 30  40 and the last class is 90  100.
Also, consider the imagined classes 20  30 and 100  110 each with frequency 0. The class marks of these classes are 25 and 105 at points A and I, respectively.
Hence, the curve ABCDEFGHI represents frequency polygon.
Page No 149:
Question 8:
Following table shows a frequency distribution for the speed of cars passing through at a particular spot on a high way :
Class interval (km/h)  Frequency 
30 – 40  3 
40 – 50  6 
50 – 60  25 
60 – 70  65 
70 – 80  50 
80 – 90  28 
90 – 100  14 
Draw the frequency polygon representing the above data without drawing the histogram.
Answer:
Page No 149:
Question 9:
Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them.
Section A  Section B  
Marks  Frequency  Marks  Frequency 
0  15 15  30 30  45 45  60 60  75 75  90 
5 12 28 30 35 13 
0  15 15  30 30  45 45  60 60  75 75  90 
3 16 25 27 40 10 
Represent the marks of the students of both the sections on the same graph by two frequency polygons. What do you observe?
Answer:
Mid marks is given by $\mathrm{Mid}\mathrm{marks}=\frac{\mathrm{Lower}\mathrm{limit}+\mathrm{Upper}\mathrm{limit}}{2}$
Section A  Section B  
Marks  MidMarks  Frequency  Marks  MidMarks  Frequency 
0  15 15  30 30  45 45  60 60  75 75  90 
7.5 22.5 37.5 52.5 67.5 82.5 
5 12 28 30 35 13 
0  15 15  30 30  45 45  60 60  75 75  90 
7.5 22.5 37.5 52.5 67.5 82.5 
3 16 25 27 40 10 
Plot the class marks along the horizontal axis and frequency along the vertical axis.
Page No 149:
Question 10:
The mean of the following distribution is 50.
x  f 
10  17 
30  5a + 3 
50  32 
70  7a – 11 
90  19 
Find the value of a and hence the frequencies of 30 and 70.
Answer:
x_{i}  f_{i}  x_{i}f_{i} 
10  17  170 
30  5a + 3  150a + 90 
50  32  1600 
70  7a – 11  490a – 770 
90  19  1710 
$\begin{array}{rcl}\therefore \mathrm{Mean}& =& \frac{\Sigma {f}_{i}{x}_{i}}{\Sigma {f}_{i}}\\ \Rightarrow 50& =& \frac{170+150a+90+1600+490a770+1710}{12a+60}\end{array}$
$\Rightarrow \frac{640a+2800}{12a+60}=50\phantom{\rule{0ex}{0ex}}\Rightarrow 640a+2800=600a+3000\phantom{\rule{0ex}{0ex}}\Rightarrow a=5$
Frequency of 30 = 5a + 3 = 5(5) + 3 = 28
Frequency of 70 = 7a – 11 = 7(5) – 11 = 24
Page No 149:
Question 11:
The mean marks (out of 100) of boys and girls in an examination are 70 and 73, respectively. If the mean marks of all the students in that examination is 71, find the ratio of the number of boys to the number of girls.
Answer:
Let number of boys and girls be x and y.
Now,
Mean marks of x boys = 70
Sum of marks of x boys = 70x .....(1)
Also,
Mean marks of y girls = 73
Sum of marks of y girls = 73y .....(2)
So,
Mean marks of (x + y) students = 71
Sum of marks of (x + y) 71(x + y) .....(3)
From (1), (2) and (3)
71(x + y) = 70x + 73y
$\Rightarrow x=2y\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{y}=\frac{2}{1}$
Thus, ratio of number of boys and girls is 2 : 1.
Page No 149:
Question 12:
A total of 25 patients admitted to a hospital are tested for levels of blood sugar, (mg/dl) and the results obtained were as follows :
87  71  83  67  85 
77  69  76  65  85 
85  54  70  68  80 
73  78  68  85  73 
81  78  81  77  75 
Find mean, median and mode (mg/dl) of the above data.
Answer:
Arranging in ascending order,
54, 65, 67, 68, 69, 70, 71, 73, 73, 75, 76, 77, 77, 78, 80, 81, 81, 83, 85, 85, 85, 85, 87
Total = 25
Sum of total numbers = 1891
$\begin{array}{rcl}\therefore \mathrm{Mean}& =& \frac{1891}{25}=75.64\\ \mathrm{Median}& =& {\left(\frac{25+1}{2}\right)}^{\mathrm{th}}\mathrm{observation}\\ & =& {13}^{\mathrm{th}}\mathrm{observation}\\ & =& 77\\ & & \end{array}$
Mode = Highest frequency element = 85.
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