Find the mean of the following distribution : -Maths 9th

1 Answer

Answer :

Now, mean =(x̅) = Σfx / Σf  = 1900 / 100 = 19

Related questions

Description : 2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Last Answer : Here's ur answer...

Description : Find the mean of the following distribution : -Maths 9th

Last Answer : Now, mean =(x̅) = Σfx / Σf = 1900 / 100 = 19

Description : Obtain the mean of the following distribution and also find the mode. -Maths 9th

Last Answer : fixi/fi 270/55 Mean=4.9

Description : 2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Last Answer : Here's ur answer...

Description : Obtain the mean of the following distribution and also find the mode. -Maths 9th

Last Answer : fixi/fi 270/55 Mean=4.9

Description : Obtain the mean of the following distribution: -Maths 9th

Last Answer : Solution :- ⇒ x̅ = ∑fx/∑f ⇒ x̅ = 332/40 ⇒ x̅ = 8.05

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 : For a particular year, following is the distribution of ages (in years) of primary school teachers in a district: -Maths 9th

Last Answer : 1.First class interval is 15 - 20 and its lower limit is 15. 2.Fourth class interval is 30 - 35 Lower limit is 30 and upper limit is 35. 3.Class mark of the class 45 - 50 =( 45+50 )/ 2 ... = Upper limit of each class interval - Lower limit of each class interval ∴ Here, class size = 20 - 15 = 5

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 : 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 : 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 : 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 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 : 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 : For a particular year, following is the distribution of ages (in years) of primary school teachers in a district: -Maths 9th

Last Answer : 1.First class interval is 15 - 20 and its lower limit is 15. 2.Fourth class interval is 30 - 35 Lower limit is 30 and upper limit is 35. 3.Class mark of the class 45 - 50 =( 45+50 )/ 2 ... = Upper limit of each class interval - Lower limit of each class interval ∴ Here, class size = 20 - 15 = 5

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 : 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 : 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 : 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 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 : 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

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

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

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 : 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

Description : To draw a histogram to represent the following frequency distribution. -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 : NEED ANSWER

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

Last Answer : NEED ANSWER

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

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.

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.

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 : (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.

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 : 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 : (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.

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

Last Answer : Frequency distribution.

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 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

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.

Description : In a frequency distribution, -Maths 9th

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

Description : The frequency distribution: -Maths 9th

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

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.

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 : 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.

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:

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

Last Answer : Solution :-

Description : find the mean of first five multiples of 10 -Maths 9th

Last Answer : This is the correct one...

Description : The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Last Answer : Here's ur answer..

Description : Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Last Answer : Since mean of 20 observations is 17 Sum of the 20 observations = 17 x 20 = 340 New sum of 20 observations = 340 – 40 + 12 = 312 New mean=312 / 20 =15.6

Description : Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Last Answer : Mean of 36 observations = 12 Total of 36 observations = 36 x 12 = 432 Correct sum of 36 observations = 432 – 74 + 47 = 405 Correct mean of 36 observations = 405/ 36 =11.25

Description : If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Last Answer : x + x + 2 + x + 4 + x + 6 + x + 8 / 5 = 11 5x + 20 = 55 5x = 35 ⇒ x = 7