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NCERT exemplar Class 9 Maths solutions chapter 14 provides the basics of statistics and probability. The concepts used here will be useful in higher classes too. These NCERT exemplar Class 9 Maths chapter 14 solutions are designed in a manner to provide the students with a correct and exhaustive approach towards the problems while practicing the questions of NCERT Book for Class 9 Maths.
The NCERT exemplar Class 9 Maths chapter 14 solutions explores all the concepts of statistics and probability recommended by the CBSE for Class 9. NCERT exemplar Class 9 Maths solutions chapter 14 are prepared by the highly experienced subject experts at Careers 360.
Also read - NCERT Solutions for Class 9 Maths
Question:1
The class mark of the class 90-120 is :
Question:3
Answer : BQuestion:4
Answer: CQuestion:5
Answer : BQuestion:6
Answer : BQuestion:7
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
Question:8
A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class intervals is constructed for the following data :
The frequency of the class 310-330 is:
Question:9
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 :
The number of classes in the distribution will be :
Question:10
To draw a histogram to represent the following frequency distribution :
Class interval | |||||
Frequency |
Question:11
Question:12
Answer:
In the given data the mean of the observations x, x + 3, x + 5, x + 7, x + 10 is given as 9Question:14
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
Question:15
Let be the mean of and the mean of . If is the mean of , then is equal to
Question:16
,
Question:17
Answer : CQuestion:18
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
Question:19
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
Question:20
Answer : BQuestion:21
The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is
Question:22
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 preceding classes
Class interval | Class marks (x-coordinate) | Frequency (y-coordinate) |
10-20 | 15 | 7 |
20-30 | 25 | 8 |
30-40 | 35 | 9 |
Question:23
Median of the following numbers : 4, 4, 5, 7, 6, 7, 7, 12, 3 is
Question:24
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
Question:26
Answer : CQuestion:27
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 |
Answer:
Answer : B
Number of students with blood group B =12
Total number of students =10+13+12+5
Required probability
Therefore option (B) is correct.
Question:28
Two coins are tossed 1000 times and the outcomes are recorded as below :
Number of heads | 2 | 1 | 0 |
Frequency | 200 | 550 | 250 |
Question:29
Life time (in hours) | 300 | 500 | 700 | 900 | 1100 |
Frequency | 10 | 12 | 23 | 25 | 10 |
Question:30
Life time (in hours) | 300 | 500 | 700 | 900 | 1100 |
Frequency | 10 | 12 | 23 | 25 | 10 |
Question:1
Marks | 0-20 | 20-40 | 40-60 | 60-100 |
Number of Students | 10 | 15 | 20 | 25 |
Answer:
Answer : yesMarks | 0-20 | 20-40 | 40-60 | 60-100 |
Number of Students | 10 | 15 | 20 | 25 |
Class interval (x-axis) | Frequency (y-axis) |
0-20 | 10 |
20-40 | 15 |
40-60 | 20 |
60-100 | 25 |
Question:2
In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded:
Which ‘average’ will be a good representative of the above data and why?
Question:3
Answer: The child has not arranged the data in ascending order before finding the medianQuestion: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
Is it the correct answer and why?
Answer:
Answer : [No, as the data is not arranged in ascending order before finding the middle terms]Question:5
Answer: NoQuestion:6
The class marks of a continuous distribution are :
Is it correct to say that the last interval will be 1.55 - 1.73? Justify your answer.
Class-marks | Difference |
1.14-1.04 | 0.1 |
1.24-1.14 | 0.1 |
1.34-1.24 | 0.1 |
1.44-1.34 | 0.1 |
1.54-1.44 | 0.1 |
1.55-1.73 | 0.8 |
Question:7
Number of hours | 0-5 | 5-10 | 10-15 | 15-20 |
Frequency | 8 | 16 | 4 | 2 |
Question:8
Can the experimental probability of an event be a negative number? If not, why?
Answer: NoQuestion:9
Can the experimental probability of an event be greater than 1? Justify your answer.
Answer : NoQuestion:10
Answer : NoQuestion: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:
Given data :Blood group | Frequency distribution table |
A | 12 |
B | 8 |
AB | 4 |
O | 6 |
Total | 30 |
Question:2
Answer:
We have to make a frequency distribution of the digits 0 to 9 after the decimal point.
Hence the frequency distribution table can be represented as:
Digit | Frequency |
0 | 1 |
1 | 2 |
2 | 5 |
3 | 6 |
4 | 3 |
5 | 4 |
6 | 3 |
7 | 2 |
8 | 5 |
9 | 4 |
Total | 35 |
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:
Given data :Scores (X) | No.of students (f) |
48 | 3 |
58 | 3 |
64 | 4 |
66 | 7 |
69 | 6 |
71 | 3 |
73 | 2 |
81 | 1 |
83 | 2 |
84 | 2 |
N=33 |
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 |
Answer:
Answer: Class size is 10Question: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 |
Answer:
Given classes are :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 |
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 |
Answer:
Here, we will draw a bar graph to represent the expenditure of a family for different heads in a month as follows:Question:7
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 |
Answer:
Expenditure on Education of a country during a five year period (2002-2006), in crores of rupees, is given below:Elementry 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 |
Question:8
Letters | a | e | i | o | r | t | u |
Frequency | 75 | 125 | 80 | 70 | 80 | 95 | 75 |
Answer:
We will draw the graph with letters (a, e, i, o, r, t, u) as follows: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 |
10 | 6 | 60 |
15 | 8 | 120 |
20 | p | 20 p |
25 | 10 | 250 |
30 | 6 | 180 |
Question:10
Obtain the meaning of the following distribution:
Frequency | Variable |
4 | 4 |
8 | 6 |
14 | 8 |
11 | 10 |
3 | 12 |
4 | 4 | 16 |
6 | 8 | 48 |
8 | 14 | 112 |
10 | 11 | 110 |
12 | 3 | 36 |
Question:11
Answer: 72.2Hence, the mean of whole class is 72.2.
Question:12
Answer :Question:13
Answer : 20Question:14
The points scored by a basketball team in a series of matches are as follows:
Find the median and mode for the data.
Answer:
Answer : [median = 12 and mode = 10]Question:15
Answer:
Based on the given bar graph :Class-Interval | Frequency |
150-200 | 50 |
200-250 | 30 |
250-300 | 35 |
300-350 | 20 |
350-400 | 10 |
Total workers | 145 |
Question:16
Monthly income (in Rs) | Number of Televisions/household | ||||
0 | 1 | 2 | Above 2 | ||
<10000 | 20 | 80 | 10 | 0 | |
10000-14999 | 10 | 240 | 60 | 0 | |
15000-19999 | 0 | 380 | 120 | 30 | |
20000-24999 | 0 | 520 | 370 | 80 | |
25000 and above | 0 | 1100 | 760 | 220 |
Answer:
Here, total events =4000Question:17
Sum | Frequency |
2 | 14 |
3 | 30 |
4 | 42 |
5 | 55 |
6 | 72 |
7 | 75 |
8 | 70 |
9 | 53 |
10 | 46 |
11 | 28 |
12 | 15 |
Answer:
Question:18
Number of defective bulbs | 0 | 1 | 2 | 3 | 4 | 5 | 6 | more than 6 |
Frequency | 400 | 180 | 48 | 41 | 18 | 8 | 3 | 2 |
Answer:
Question:19
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 |
Answer:
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 |
Answer:
(iv) p(under 60 but over 39 years) = p(person having age 40 to 49 years) + p (person having age 50 to 59 years)
Question:1
The following are the marks (out of 100) of 60 students in mathematics.
Construct a grouped frequency distribution table with width 10 of each class starting from 0 - 9.
Answer:
Step 1: arrange these number in ascending order.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 | 60 |
Question:2
The following are the marks (out of 100) of 60 students in mathematics.
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:
Step 1: arrange these number in ascending order.Class interval | Frequency |
0-10 | 4 |
10-20 | 7 |
20-30 | 5 |
30-40 | 10 |
40-50 | 5 |
50-60 | 8 |
60-70 | 5 |
70-80 | 8 |
80-90 | 5 |
90-100 | 3 |
Total | 60 |
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:
To draw the histogram, we have to plot the frequency, i.e., number of students on the y-axis and class interval, i.e., heights on the x-axis. We can construct the histogram as follows: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:
We can see that the intervals are 20-24, 25-29, 30-34....etcAges | No. 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 |
Question:5
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 |
Answer:
We can see that the intervals are 118-126, 127-135, 136-144...etcLength (in mm) | Number of leaves |
117.5-126.5 | 8 |
126.5-135.5 | 10 |
135.5-144.5 | 12 |
144.5-153.5 | 17 |
153.5-162.5 | 7 |
162.5-171.5 | 5 |
171.5-180.5 | 3 |
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 |
Answer:
To draw the histogram, we have to plot the frequency, i.e., number of students on the y-axis and class interval, i.e., marks on the x-axis. We can construct the histogram as follows:Question:7
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 |
Answer:
To draw the histogram, we have to plot the frequency on the y-axis and class interval on the x-axis. We can construct the histogram as follows:Question:8
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 |
Answer:
A frequency polygon is a graph constructed by using lines to join the midpoints of each interval.Question:9
Section A | Section B | ||
Marks | Frequency | Marks | Frequency |
0-15 | 5 | 0-15 | 3 |
15-30 | 12 | 15-30 | 16 |
30-45 | 28 | 30-45 | 25 |
45-60 | 30 | 45-60 | 27 |
60-75 | 35 | 60-75 | 40 |
75-90 | 13 | 75-90 | 10 |
Answer:
Marks | Class marks | Frequency A | Frequency B |
0-15 | 5 | 3 | |
15-30 | 12 | 16 | |
30-45 | 28 | 25 | |
45-60 | 30 | 27 | |
60-75 | 35 | 40 | |
75-90 | 13 | 10 |
Question:10
The mean of the following distribution is 50.
Find the value of a and hence the frequencies of 30 and 70.Answer:
10 | 17 | 170 |
30 | ||
50 | 32 | 1600 |
70 | ||
90 | 19 | |
Total |
Question:10
The mean of the following distribution is 50.
x | f |
10 | |
30 | |
50 | |
70 | |
90 |
Answer:
10 | 17 | 170 |
30 | ||
50 | 32 | 1600 |
70 | ||
90 | 19 | |
Total |
Question:11
Answer : [ 2:1 ]Question:12
Answer:
Answer : The mean is 75.88, mode is 85, median is 77Mode is the value that occurs the most number of times in a given set of values
So, 85 is the mode, and its frequency is 4 times.
To calculate the median, arrange the given data in ascending order and then find the middle term. This middle term is called the median.
The data in ascending order is:
Number of terms, n=25
13^{th} term is 77
Hence, median is 77
The important topics covered in NCERT exemplar Class 9 Maths solutions chapter 14 are mentioned below:
These Class 9 Maths NCERT exemplar chapter 14 solutions provide a basic knowledge of statistics and probability which has great importance in economics, politics, social science as well as in science. These solutions can be used as supporting material to better study Statistics and Probability-based practice problems. The Class 9 Maths NCERT exemplar solutions chapter 14 Statistics and Probability provides adequate practice and information on this chapter in a student-friendly process flow that can be further applied while solving other books such as Class 9 Mathematics by Pearson, NCERT Class 9 Maths, RD Sharma Class 9 Maths, RS Aggarwal Class 9 Maths etcetera.
NCERT exemplar Class 9 Maths solutions chapter 14 pdf download is a free to use feature for students and can be used judiciously while studying NCERT exemplar Class 9 Maths chapter 14 in an offline environment.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
No, the possibility of occurring or not occurring may have different methods. Therefore, we can say occurring or non-occurring are not equally likely. If two events are not equally likely their probability will not be the same.
Mean, median, and mode of any data represent the central tendency of the data.
Mode represents the most favourable data outcome.
Mean represent the average value of the data.
Median represents the middle value, if the data is arranged in decreasing or increasing order.
When the occurrence of any event cannot be denied, the event is called a sure event. For example, the sun will rise in the east can be seen as a sure event. The probability of a sure event is one.
The probability of any event cannot be more than one. The maximum probability is 1 for sure events. The minimum probability is 0 for impossible events
On the basis of analysis of large data, we estimate the occurrence of any event this analysis is called statistical probability. Most of the recent scientific researches are based on statistical probability
The chapter on Statistics and Probability is a very important chapter from the exam’s perspective as it accounts for 10-12% marks of the final paper. NCERT exemplar Class 9 Maths solutions chapter 14 provides sufficient learning content to the students to develop a better understanding and score well in the exam.
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