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

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

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.

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.

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.

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.

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

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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)

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 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

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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.

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 -Maths 9th

Description : 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 -Maths 9th

Last Answer : NEED ANSWER

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Description : 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. -Maths 9th

Last Answer : NEED ANSWER

If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Description : If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Last Answer : NEED ANSWER

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, -Maths 9th

Description : 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, -Maths 9th

Last Answer : NEED ANSWER

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 -Maths 9th

Description : 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 -Maths 9th

Last Answer : (b) Let x and y be the upper and lower class limit in a frequency distribution. Now, mid value of a class (x + y )/2=10 [given] ⇒ x + y = 20 (i) Also, given that, width of class x- y = 6 (ii) On ... putting x = 13 in Eq. (i), we get 13+y = 20 ⇒ y = 7 Hence, the lower limit of the class is 7.

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Description : 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. -Maths 9th

Last Answer : (c) Let x and y be the upper and lower class limit of frequency distribution. Given, width of the class = 5 ⇒ x-y= 5 (i) Also, given lower class (y) = 10 On putting y = 10 in Eq. (i), we get ... 20-25, 25-30 and 30-35. Thus, the highest class is 30-35, Hence, the upper limit of this class is 35.

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Last Answer : (b) Let x and y be the lower and upper class limit of a continuous frequency distribution. Now, mid-point of a class = (x + y)/2 = m [given] ⇒ x + y = 2 m =x + l = 2m [∴ y = l = upper class limit (given)] ⇒ x = 2 m-l Hence, the lower class limit of the class is 2m – l.

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, -Maths 9th

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Last Answer : (b) We arrange the given data into groups like 13-22,23-32 103-112. (since, our data is from 14 to 112). The class width in this case is 9. Now, the given data can be arranged in tabular form as follows. Hence, the number of classes in distribution will be 10.

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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.

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

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

Last Answer : NEED ANSWER

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

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

The frequency distribution has been represented graphically as follows. -Maths 9th

Description : The frequency distribution has been represented graphically as follows. -Maths 9th

Last Answer : NEED ANSWER

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

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

Last Answer : Frequency distribution.

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

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.

The frequency distribution has been represented graphically as follows. -Maths 9th

Description : The frequency distribution has been represented graphically as follows. -Maths 9th

Last Answer : No, here the widths of the rectangles are varying, so we need to make certain modifications in the length of the rectangles so that the areas are proportional to the frequencies. We proceed as ... Now, we get the following modified table So, the correct histogram with varying width is given below

In a frequency distribution, -Maths 9th

Description : In a frequency distribution, -Maths 9th

Last Answer : Lower limit of the class = 10 - 1/2 x 6 = 10 - 3 = 7

The frequency distribution: -Maths 9th

Description : The frequency distribution: -Maths 9th

Last Answer : No, as the classes are of varying widths, not of uniform widths.

Convert the given frequency distribution into a continuous -Maths 9th

Description : Convert the given frequency distribution into a continuous -Maths 9th

Last Answer : Consider the classes 150 - 153 and 154 - 157. The lower limit of 154 - 157 = 154 The upper limit of 150 - 153 = 153 The difference = 154 - 153 = 1 Half the difference = 1/2 = 0.5 So, the ... Continuous classes formed are: 153.5 is included in the class interval 153.5 - 157.5 and 157.5 - 161.5.

Prepare a continuous grouped frequency distribution from the following data: -Maths 9th

Description : Prepare a continuous grouped frequency distribution from the following data: -Maths 9th

Last Answer : If m is mid-point of a class and h is the class size, lower and upper limits of the class intervals are m - h/2 and m + h/2 respectively. Class size (h) = 15 - 5 = 10 So, the class interval formed ... 2) - (5 + 10/2) i.e., 0 - 10 Continuing in the same manner, the continuous classes formed are:

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

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

Last Answer : Solution :-

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!!!

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

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

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)

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!!!

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

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

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)

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 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