Class 9 Mathematics · 62 questions · 30 with answers
The marks obtained by 17 students in a mathematics test (out of 100) are given below : 91, 82, 100, 100, 96, 65, 82, 76, 79, 90, 46, 64, 72, 68, 66, 48, 49. The range of the data is :
Solution : Answer (B)
The class-mark of the class 130-150 is :
Solution : Answer (C)
A die is thrown 1000 times and the outcomes were recorded as follows : Outcome 1 2 3 4 5 6 Frequency 180 150 160 170 150 190 If the die is thrown once more, then the probability that it shows 5 is :
The mean of the data : 2, 8, 6, 5, 4, 5, 6, 3, 6, 4, 9, 1, 5, 6, 5 is given to be 5. Based on this information, is it correct to say that the mean of the data: 10, 12, 10, 2, 18, 8, 12, 6, 12, 10, 8, 10, 12, 16, 4 is 10? Give reason.
Solution : It is correct. Since the 2nd data is obtained by multiplying each observation of 1st data by 2, therefore, the mean will be 2 times the mean of the 1st data.
In a histogram, the areas of the rectangles are proportional to the frequencies. Can we say that the lengths of the rectangles are also proportional to the frequencies?
Solution: No. It is true only when the class sizes are the same. Sample Quetion 3 : Consider the data : 2, 3, 9, 16, 9, 3, 9. Since 16 is the highest value in the observations, is it correct to say that it is the mode of the data? Give reason. Solution : 16 is not the mode of the data. The mode of a given data is the observation with highest frequency and not the observation with highest value.
Heights (in cm) of 30 girls of Class IX are given below: 140, 140, 160, 139, 153, 153, 146, 150, 148, 150, 152, 146, 154, 150, 160, 148, 150, 148, 140, 148, 153, 138, 152, 150, 148, 138, 152, 140, 146, 148. Prepare a frequency distribution table for this data.
Solution : Frequency distribution of heights of 30 girls Height Tally Marks Frequency (in cm) 138 || 2 139 | 1 140 |||| 4 146 ||| 3 148 |||| | 6 150 |||| 5 152 ||| 3 153 ||| 3 154 | 1 160 || 2 Total 30 STATISTICS AND PROBABILITY 139
The following observations are arranged in ascending order : 26, 29, 42, 53, x, x + 2, 70, 75, 82, 93 If the median is 65, find the value of x.
Solution : Number of observations (n) = 10, which is even. Therefore, median is the n n mean of and + 1 observation, i.e., 5th and 6th observation. 2 2 Here, 5th observation = x 6th observation = x + 2 x + ( x + 2) Median = = x +1 Now, x + 1 = 65 (Given) Therefore, x = 64 Thus, the value of x is 64.
Here is an extract from a mortality table. Age (in years) Number of persons surviving out of a sample of one million 60 16090 61 11490 62 8012 63 5448 64 3607 65 2320
Solution : (i) We see that 16090 persons aged 60, (16090-11490), i.e., 4600 died before reaching their 61st birthday. 4600 460 Therefore, P(a person aged 60 die within a year) = = 16090 1609 (ii) Number of persons aged 61 years = 11490 Number of persons surviving for 4 years = 2320 2320 232 P(a person aged 61 will live for 4 years) = = 11490 1149
Following is the frequency distribution of total marks obtained by the students of different sections of Class VIII. Marks 100 - 150 150 - 200 200 - 300 300 - 500 500 - 800 Number of students 60 100 100 80 180 Draw a histogram for the distribution above.
Solution: In the given frequency distribution, the class intervals are not of equal width. Therefore, we would make modifications in the lengths of the rectangles in the histogram so that the areas of rectangles are proportional to the frequencies. Thus, we have: Marks Frequency Width of the class Length of the rectangle 100 - 150 60 50 × 60 = 60 150 - 200 100 50 × 100 = 100 200 - 300 100 100 × 100 = 50 300 - 500 80 200 × 80 = 20 500 - 800 180 300 × 180 = 30 Now, we draw rectangles with lengths as given in the last column. The histogram of the data is given below : Fig. 14.3
Two sections of Class IX having 30 students each appeared for mathematics olympiad. The marks obtained by them are shown below:
3 4 7
The class mark of the class 90-120 is :
(B) 105
The range of the data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is
(D) 26
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 :
(B) 7
The width of each of five continuous classes in a frequency distribution is 5 and the lower class-limit of the lowest class is 10. The upper class-limit of the highest class is:
(C) 35
Let m be the mid-point and l be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is :
(B) 2m – l
The class marks of a frequency distribution are given as follows : 15, 20, 25, ... The class corresponding to the class mark 20 is :
(B) 17.5 – 22.5
In the class intervals 10-20, 20-30, the number 20 is included in :
(B) 20-30
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 310-330 is:
(C) 6
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 :
(B) 10
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 STATISTICS AND PROBABILITY 133 the adjusted frequency for the class 25-45 is :
(D) 2
The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is :
(D) 38
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
(C)
2 1 2
(B) 10
3 3 3 13. If x represents the mean of n observations x1, x2, ..., xn, then value of ( xi − x ) is: i =1
(B) 0
13 3 1
(B)
40 10 8 28. 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
(C)
1 4 3
(B)
7
(B)
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 : STATISTICS AND PROBABILITY 137 Fig. 14.1 Do you think this representation is correct? Why?
This question refers to a figure in the original PDF.
Not correct. The classes are of varying widths, not of uniform widths.
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?
Median will be a good representative of the data, because (i) each value occcurs once, (ii) The data is influenced by extreme values.
A child says that the median of 3, 14, 18, 20, 5 is 18. What doesn’t the child understand about finding the median?
Data has to be arranged in ascending (or descending ) order before finding the median.
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 5th observation + 6 th observation = 8+6 = =7 Is it the correct answer and why?
No, the data have first to be arranged in ascending (or descending) order before finding the median.
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.
It is not correct. In a histogram, the area of each rectangle is propotional to the frequency of its class.
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.
It is not correct. Reason is that differnce between two consecutive marks should be equal to the class size.
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.
No. Infact the number of children who watch TV for 10 or more hours a week is 4 + 2, i.e., 6.
Can the experimental probability of an event be a negative number? If not, why?
No, since the number of trials in which the event can happen cannot be negative, and the total number of trials is always positive.
Can the experimental probability of an event be greater than 1? Justify your anwer.
No, since the number of trials in which the event can happen cannot be greater than the total number of trials.
As the number of tosses of a coin increases, the ratio of the number of heads to the total number of tosses will be . Is it correct? If not, write the correct one.
No. As the number of tosses of a coin increases, the ratio of the number of 1 1 heads to the total number of tosses will be nearer to , not exactly . 2 2 EX ERCISE 14.3 1. Blood Group Number of Students (frequency) A 12 B 8 AB 4 O 6 Total 30 2. Digit 0 1 2 3 4 5 6 7 8 9 Frequency 1 2 5 6 3 4 3 2 5 4 ANSWERS 167 3. Scores 48 58 64 66 69 71 73 81 83 84 Frequency 3 3 4 7 6 3 2 1 2 2 4. Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 4 8 13 12 6 Class size = 10 5. Class intervals 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 6 153.5 is included in the class interval 153.5-157.5 and 157.5 in 157.5 - 161.5. 9. 20 10. 8.05
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.
The value of π upto 35 decimal places is given below:
14159265358979323846264338327950288 Make a frequency distribution of the digits 0 to 9 after the decimal point. 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, 84 000 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.
Prepare a continuous grouped frequency distribution from the following data: Mid-point Frequency
4 15 8 25 13 35 12 45 6 Also find the size of class intervals. 5. Convert the given frequency distribution into a continuous grouped frequency distribution: STATISTICS AND PROBABILITY 141 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?
The expenditure of a family on different heads in a month is given below: Head Food Education Clothing House Rent Others Savings Expenditure 4000 2500 1000 3500 2500 1500 (in Rs) Draw a bar graph to represent the data above.
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.
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.
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
Obtain the mean of the following distribution: Frequency Variable 4 4 8 6 14 8
10 3 12 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.
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.
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.
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.
In Fig. 14.2, there is a histogram depicting daily wages of workers in a factory. Construct the frequency distribution table. Fig. 14.2 STATISTICS AND PROBABILITY 143
This question refers to a figure in the original PDF.
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 Number of Televisions/household (in Rs) 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 Find the probability:
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
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
Over the past 200 working days, the number of defective parts produced by a machine is given in the following table: Number of 0 1 2 3 4 5 6 7 8 9 10 11 12 13 defective parts Days 50 32 22 18 12 12 10 10 10 8 6 6 2 2 Determine the probability that tomorrow’s output will have
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:
31 74 68 42 54 14 61 83 48 37 26 8 64 57 93 72 53 59 38 16 88 75 56 46 66 45 61 54 27 27 44 63 58 43 81 64 67 36 49 50 76 38 47 55 77 62 53 40 71 60 58 45 42 34 46 40 59 42 29 Construct a group frequency distribution of the data above using the classes 0-9, 10-19 etc., and hence find the number of students who secured more than 49 marks. Solution : Class Tally Marks Frequency 0-9 | 1 10-19 || 2 20-29 |||| 4 30-39 |||| | 6 40-49 |||| |||| |||| 15 50-59 |||| |||| || 12 60-69 |||| |||| 10 70-79 |||| | 6 80-89 ||| 3 90-99 | 1 Total 60 STATISTICS AND PROBABILITY 147 From the table above, we find that the number of students who secure more than 49 marks is (12 + 10 + 6 + 3 + 1), i.e., 32.
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.
Class 0 - 9 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 80 - 89 90 - 99 Frequency 1 2 5 6 3 4 3 2 5 4
Refer to Q1 above. 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).
Class intervals 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 10. a = 5, frequency of 30 is 28 and that of 70 is 24. 11. 2 : 1 12. Mean = 75.64, Median = 77, Mode = 85
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
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
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.
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.
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. STATISTICS AND PROBABILITY 149
Refer to Q. 7 : Draw the frequency polygon representing the above data without drawing the histogram.
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 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 Represent the marks of the students of both the sections on the same graph by two frequency polygons.What do you observe?
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.
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.
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.