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

1 Answer

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 .

Related questions

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 .

Description : Draw a histogram and frequency polygon for the following distribution : -Maths 9th

Last Answer : We represent class limits along x-axis and number of students along y-axis on a suitable Scale.

Description : Draw a histogram and frequency polygon for the following distribution : -Maths 9th

Last Answer : We represent class limits along x-axis and number of students along y-axis on a suitable Scale.

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!!!

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!!!

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

Last Answer : 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 ... draw rectangles with lengths as given in the last column. The histogram of data is given below:

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

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

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.

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.

Description : To draw a histogram to represent the following frequency distribution : -Maths 9th

Last Answer : Adjusted frequency of a class = Minimum class size of frequency distribution × Frequency of given class / Class size of given class ∴ Adjusted frequency for the class 25 - 45 = 5 × 8 / 20 = 2

Description : To draw a histogram to represent the following frequency distribution : -Maths 9th

Last Answer : Adjusted frequency of a class = Minimum class size of frequency distribution × Frequency of given class / Class size of given class ∴ Adjusted frequency for the class 25 - 45 = 5 × 8 / 20 = 2

Description : To draw a histogram to represent the following frequency distribution. -Maths 9th

Last Answer : NEED ANSWER

Description : To draw a histogram to represent the following frequency distribution. -Maths 9th

Last Answer : Frequency distribution.

Description : In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Last Answer : Frequency distribution table

Description : In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Last Answer : Frequency distribution table

Description : The following table gives the pocket money (in Rs) given to children per day by their parents : Represent the data in the form of a histogram. -Maths 9th

Last Answer : The required histogram is as below :

Description : The following table gives the pocket money (in Rs) given to children per day by their parents : Represent the data in the form of a histogram. -Maths 9th

Last Answer : The required histogram is as below :

Description : Draw a frequency polygon for the following distribution: -Maths 9th

Last Answer : Solution :-

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)

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)

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

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

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

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

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.

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

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

Description : 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 -Maths 9th

Last Answer : NEED ANSWER

Description : 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 -Maths 9th

Last Answer : (c) Class marks i.e., the mid-point of the classes are abscissa of the points, which we plot for frequency polygon.

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

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)

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.

Description : 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. -Maths 9th

Last Answer : It is not correct. Because in a histogram, the area of each rectangle is proportional to the corresponding frequency of its class.

Description : 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. -Maths 9th

Last Answer : It is not correct. Because in a histogram, the area of each rectangle is proportional to the corresponding frequency of its class.

Description : In a histogram, the areas of the rectangles are proportional -Maths 9th

Last Answer : No. It is true only when the class sizes are the same.

Description : Is it correct to say that in a histogram, -Maths 9th

Last Answer : It is not correct. In a histogram, the area of each rectangle is proportional to the frequency of its class.

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.

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.

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.

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.

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

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.

Description : How many diagonals can be drawn in a polygon of 15 sides? -Maths 9th

Last Answer : answer:

Description : What do you mean by polygon? -Maths 9th

Last Answer : answer:

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

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

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

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

Description : 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 -Maths 9th

Last Answer : NEED ANSWER