The marks obtained (out of 100) by a class of 80 students are given below: -Maths 9th

The marks obtained (out of 100) by a class of 80 students are given below: -Maths 9th
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In the given frequency distribution, the class intervals are not of equal width. Therefore, we would make modification in the lengths of the rectangle in the histogram so that the areas of rectangle are proportional to the frequencies. Thus we have:  Now, we draw rectangles with lengths as given in the last column. The histogram of data is given below:
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Related questions

In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Description : In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Last Answer : Here, class size = 135 - 305 = 10 ∴ Lower limit of first class interval is 305 - 10 / 2 = 300 Upper limit of first class interval is 305 + 10 / 2 = 310 Thus, first class interval is 300 - 310 Required histogram and frequency polygon is given on the graph paper .

In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Description : In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Last Answer : Here, class size = 135 - 305 = 10 ∴ Lower limit of first class interval is 305 - 10 / 2 = 300 Upper limit of first class interval is 305 + 10 / 2 = 310 Thus, first class interval is 300 - 310 Required histogram and frequency polygon is given on the graph paper .

Following is the frequency distribution of total marks obtained by the students of different section of class-IX. -Maths 9th

Description : Following is the frequency distribution of total marks obtained by the students of different section of class-IX. -Maths 9th

Last Answer : Since class intervals of the given frequency distribution are not of equal width. We would make modifications in the lengths of the rectangles in the histogram, so that the areas of rectangles are proportional ... lengths as given in the last column .The histogram of the data is given below.

Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them. -Maths 9th

Description : Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them. -Maths 9th

Last Answer : Clearly, the mean score of two sections A and B is same

Following is the frequency distribution of total marks obtained by the students of different section of class-IX. -Maths 9th

Description : Following is the frequency distribution of total marks obtained by the students of different section of class-IX. -Maths 9th

Last Answer : Since class intervals of the given frequency distribution are not of equal width. We would make modifications in the lengths of the rectangles in the histogram, so that the areas of rectangles are proportional ... lengths as given in the last column .The histogram of the data is given below.

Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them. -Maths 9th

Description : Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them. -Maths 9th

Last Answer : Clearly, the mean score of two sections A and B is same

The percentage of marks obtained by a student in monthly unit tests are given below. -Maths 9th

Description : The percentage of marks obtained by a student in monthly unit tests are given below. -Maths 9th

Last Answer : (i) Number of tests in which the student scored more than 70% marks = 3 ∴ P(more than 70% marks) = 3/6 = 1/2 (ii) Number of tests in which the student scored less than 70% marks = 3 ∴ P(less ... ) Number of tests in which the student scored at least 60% marks = 5 ∴ P(at least 60% marks) = 5/6

Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Description : Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Last Answer : see in book okay!!!

Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Description : Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Last Answer : see in book okay!!!

In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

Description : In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

Last Answer : For Mean : As we know that ⇒ x̅ = 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60 / 15 x̅ = 822 / 15 = 54.8 ⇒ x̅ = 54.8 For Median : First of ... 52 For Mode : Make a frequency table for given data : Here, the marks 52 has the maximum frequency '3'. ∴ Mode = 52

In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

Description : In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

Last Answer : For Mean : As we know that ⇒ x̅ = 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60 / 15 x̅ = 822 / 15 = 54.8 ⇒ x̅ = 54.8 For Median : First of ... 52 For Mode : Make a frequency table for given data : Here, the marks 52 has the maximum frequency '3'. ∴ Mode = 52

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 and 44. -Maths 9th

Description : 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 and 44. -Maths 9th

Last Answer : NEED ANSWER

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 and 44. -Maths 9th

Description : 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 and 44. -Maths 9th

Last Answer : Median will be a good representative of the data, because 1.each value occurs once. 2.the data is influenced by extreme values.

The number of students in four classes A, B, C, D and their respective mean marks obtained by each of the class are given below : Image The combined mean of the marks of four classes together will be : (A) 32 (B) 50 (C) 20 (D) 15

Description : The number of students in four classes A, B, C, D and their respective mean marks obtained by each of the class are given below :   The combined mean of the marks of four classes together will be : (A) 32 (B) 50 (C) 20 (D) 15 

Last Answer : Answer: A

The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, 48, 29, 40, 45, 48, 42, 23, 35. Find the average mark

Description : The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, 48, 29, 40, 45, 48, 42, 23, 35. Find the average mark

Last Answer : The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, ... , 23, 35. Find the average mark of the students.

The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Description : The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Last Answer : Median=n=even =n/2=20/2=10th observation =98 Mode =88

The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Description : The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Last Answer : Median=n=even =n/2=20/2=10th observation =98 Mode =88

A class consists of 80 students, 25 of them are girls and 55 boys. 10 of them are rich and 20 are fair complexioned. -Maths 9th

Description : A class consists of 80 students, 25 of them are girls and 55 boys. 10 of them are rich and 20 are fair complexioned. -Maths 9th

Last Answer : Let P (A) = Probability of selecting a fair complexioned person. ThenP(A) = \(rac{20}{80}\) = \(rac{1}{4}\)Let P(B) = Probability of selecting a rich person. Then P(B) = \(rac{10}{80}\) = \(rac{1}{8}\)Let P (C) = ... ) = \(rac{1}{4}\)x \(rac{1}{8}\)x \(rac{5}{16}\) = \(rac{5}{512}\) = 0.009.

The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Description : The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Last Answer : NEED ANSWER

The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Description : The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Last Answer : (b) Since, the difference between mid value is 5. So, the corresponding class to the class mark 20 must have difference 5. Therefore, option (c) and (d) are wrong. Since, the mid value is 20 which can get only, if we take option (b)

The table shows the marks obtained by a student in unit tests out of 50 : -Maths 9th

Description : The table shows the marks obtained by a student in unit tests out of 50 : -Maths 9th

Last Answer : Here the marks are out of 50 , so we find its percentage (i.e. out of 100)

The table shows the marks obtained by a student in unit tests out of 50 : -Maths 9th

Description : The table shows the marks obtained by a student in unit tests out of 50 : -Maths 9th

Last Answer : Here the marks are out of 50 , so we find its percentage (i.e. out of 100)

The average marks of a class of 90 students is 126. Out Of them, 4 scores zero, first 60 students scored an average of 116, next 24 scored an average of 118. What is the mark obtained by the remaining student in the class? A) 750 B) 862 C) 774 D) 875

Description : The average marks of a class of 90 students is 126. Out Of them, 4 scores zero, first 60 students scored an average of 116, next 24 scored an average of 118. What is the mark obtained by the remaining student in the class? A) 750 B) 862 C) 774 D) 875

Last Answer : C) According to the question, 90*126=(4*0)+(60*116)+(24*118)+2y 11340=6960+2832+2y 2y=1548 y=774

The mean marks obtained by a class of 40 students is 65. The mean marks of half of the students is found to be 45. The mean marks of the remaining students is (A) 85 (B) 60 (C) 70 (D) 65

Description : The mean marks obtained by a class of 40 students is 65. The mean marks of half of the students is found to be 45. The mean marks of the remaining students is (A) 85 (B) 60 (C) 70 (D) 65

Last Answer : Answer: A

The average of marks obtained by 60 students in a computer examination is 18. If the average marks of passed students is 20and that of the failed students is 8, what is the number of students who passed the examination? A) 100 B) 75 C) 50 D) 85

Description : The average of marks obtained by 60 students in a computer examination is 18. If the average marks of passed students is 20and that of the failed students is 8, what is the number of students who passed the examination? A) 100 B) 75 C) 50 D) 85 

Last Answer : C Let the number of passed students be x. Then total marks = 60 × 18= 20x + (60– x) × 8 1080= 20x + 480– 8x 12x = 600 ∴ x = 50 ∴ number of passed students = 50

The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.

The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.

The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Description : The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Last Answer : NEED ANSWER

The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Description : The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Last Answer : It is not correct. Because the difference between two consecutive class marks should be equal to the class size. Here, difference between two consecutive marks is 0.1 and class size of 1.55-1.73 is 0.18, which are not equal.

The class marks of a continuous distribution are: -Maths 9th

Description : The class marks of a continuous distribution are: -Maths 9th

Last Answer : It is not correct because the difference between two consecutive marks should be equal to the class size.

If the class marks in frequency distribution are -Maths 9th

Description : If the class marks in frequency distribution are -Maths 9th

Last Answer : The class size of the distribution is = 40.5 - 33.5 = 7 The required class of the class mark 33.5 is [33.5 - 7/2] - [33.5 + 7/2], i.e., 30 - 37.

The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Description : The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Last Answer : (a) P (later half of the year) = 23 / 40 (b) P (month having 31 days) = 26 / 40 = 13 / 20 (c) P(month having 30 days) = 10 / 40 = 1 / 4

The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Description : The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Last Answer : (a) P (later half of the year) = 23 / 40 (b) P (month having 31 days) = 26 / 40 = 13 / 20 (c) P(month having 30 days) = 10 / 40 = 1 / 4

What are the characteristics of Continuous and Comprehensive Evaluation ? (a) It increases the workload on students by taking multiple tests. (b) It replaces marks with grades. (c) It evaluates every aspect of the student. (d) It helps in reducing examination phobia. Select the correct answer from the codes given below: (A) (a), (b), (c) and (d) (B) (b) and (d) (C) (a), (b) and (c) (D) (b), (c) and (d)

Description : What are the characteristics of Continuous and Comprehensive Evaluation ? (a) It increases the workload on students by taking multiple tests. (b) It replaces marks with grades. (c) It evaluates every aspect of the student. (d) It helps in ... ) and (d) (C) (a), (b) and (c) (D) (b), (c) and (d)

Last Answer : Answer: D

In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately. -Maths 10th

Description : In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately. -Maths 10th

Last Answer : Let P has obtained x in mathematics and y in science. Then by the problem, x+y=28.......(1). If P would have got 3 marks more in mathematics, then P would have got (x+3) in mathematics ... P obtained 12 marks in mathematics and 16 in science. or, P obtained 9 marks in mathematics and 19 in science.

A class consists of 50 students out of which 30 are girls. -Maths 9th

Description : A class consists of 50 students out of which 30 are girls. -Maths 9th

Last Answer : Mean marks scored by girls ( x̅1 ) = 73 Number of girls (n1) = 30 Mean marks scored by boys ( x̅2 ) = 71 Number of boys (n2) = 50 - 30 = 20 Mean score of the whole class ( x̅12 ) = n1 x̅1 + n2 x̅2 /n1 + n2 = 30 x 73 + 20 x 71/30 + 20 = 2190 + 1420/50 = 3610/50 x̅ 2 = 72.2

Two students appeared at an examination. One of them secured 18marks more than the other and his marks was 72% of the sum of their marks. What are the marks obtained by them? a) 12.5,23.3 b) 26.7,16.0 c) 13.3,14.2 d) 11.45, 29.45 e) 29.8,15.4

Description : Two students appeared at an examination. One of them secured 18marks more than the other and his marks was 72% of the sum of their marks. What are the marks obtained by them? a) 12.5,23.3 b) 26.7,16.0 c) 13.3,14.2 d) 11.45, 29.45 e) 29.8,15.4 

Last Answer : Answer: D  Let the marks secured by them be x and (x + 18)  Then sum of their marks = x + (x + 18) = 2x + 18  Given that (x + 18) was 72% of the sum of their marks  =>(x+18) = 72/100(2x+18)  => ... 11x = 126 x = 11.45  Then (x + 18) = 11.45 + 18 = 29.45  Hence their marks are 11.45 and 29.45

In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Description : In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 Number of students having blood group ‘B’ = 12 Required probability =12 / 40 = 3 / 10

In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Description : In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 Number of students having blood group ‘B’ = 12 Required probability =12 / 40 = 3 / 10

The ratio of girls and boys in a class is 1: 3. Set up an equation between the students of a class and boys and then draw its graph. Also find the number of boys in a class of 40 students from the graph. -Maths 9th

Description : The ratio of girls and boys in a class is 1: 3. Set up an equation between the students of a class and boys and then draw its graph. Also find the number of boys in a class of 40 students from the graph. -Maths 9th

Last Answer : Total number of boys and girl = 40, Ratio = 1 : 3 Number of girls be A and Number of boys be B. Ratio of number of girls and boys is 1 : 3, so Therefore 3A=B To find number of boys we ... the number 30 represents the number of girls. 40 as total on the line A = 10, which is the common equation.

The blood groups of 30 students of class VIII are recorded as follows: -Maths 9th

Description : The blood groups of 30 students of class VIII are recorded as follows: -Maths 9th

Last Answer : Frequency Distribution Table Blood group O is most common as it has highest frequency, i.e., 12. Blood group AB is rarest as it has lowest frequency, i.e., 3.

In a medical examination of students of a class, -Maths 9th

Description : In a medical examination of students of a class, -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 P (a student has blood group B) = 12/40 = 3/10

The number of sincere students (x) in a class -Maths 9th

Description : The number of sincere students (x) in a class -Maths 9th

Last Answer : x = 2 + 2y Sincere students always progress in life as they value time and channelise their talent in productive activities while a careless student always wastes his talent and time.

Consider the table STUDENT given below: RollNo Name Class DOB Gender City Marks 1 Anand XI 6/6/97 M Agra 430 -Technology

Description : Consider the table STUDENT given below: RollNo Name Class DOB Gender City Marks 1 Anand XI 6/6/97 M Agra 430 -Technology

Last Answer : (1)b. Both (iii) and (iv)(2) b.RollnoNameClassDOBGenderCityMarks6734MaakhiyNehaGeetPreetiXIXXIXII12/12/948/12/956/5/978/8/95FFFFDubaiMoscowAgraMumbai256324470492

how i can score more marks in sa1 -Maths 9th

Description : how i can score more marks in sa1 -Maths 9th

Last Answer : u need to focus on ncert book than iinterfaces onlin if ur weak in thes u should do these math example science pratical social explanation surf internet on ur topic hindi open up with others inspeeaking english same

how i can score more marks in sa1 -Maths 9th

Description : how i can score more marks in sa1 -Maths 9th

Last Answer : u need to focus on ncert book than iinterfaces onlin if ur weak in thes u should do these math example science pratical social explanation surf internet on ur topic hindi open up with others inspeeaking english same

Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5