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The given data is not continous we therefore modify the limits of the class intervals as well to make the class intervals continous. To make the frequency polygon we first modify the table as follows To make the frequency polygon we mark the number of balls on the x-axis and the runs scored on the y-axis. The representation of the given information in the form of frequency polygon is as follows.
The representation of the given data in the form of a frequency distribution table is as follows.

D Divya Prakash Singh
First, we make  and then its bisector. The steps of constructions are: 1. Draw a ray OA. 2. Taking O as the center and any radius of your own choice, draw an arc cutting OA at B. 3. Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C. 4. Draw the ray OD passing through the C. Thus,  Now, we draw bisector of  5. Taking C and...
The mean is not an appropriate measure of central tendency in case the observations are not close to each other. An example of such a case is prices of the toys in a toy shop.
The mean is an appropriate measure of central tendency in case the observations are close to each other. An example of such a case is height of the students in a class.
Salary ( in Rs)(xi) Number of workers(fi) fixi 3000 16 48000 4000 12 48000 5000 10 50000 6000 8 48000 7000 6 42000 8000 4 32000 9000 3 27000 10000 1 10000 Total   The mean of the above data is given by The mean salary of the workers working in the factory is Rs 5083.33
In the given data 14 is occuring the maximum number of times (4) Mode of the given data is therefore 14.
The given data is already in ascending order Number of observations, n = 10 (even) x + 1 = 63 x = 62
Number of observations, n = 15 Mean is 54.8 To find the median we have to arrange the given data in ascending order as follows: 39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98 n = 15 (odd) In the given data 52 occurs the maximum number of times () Therefore, Mode = 52
Number of observations, n = 10 Mean is 2.8 To find the median we have to arrange the given data in ascending order as follows: 0, 1, 2, 3, 3, 3, 3, 4, 4, 5 n = 10 (even) In the given data 3 occurs the maximum number of times (4) Therefore, Mode = 3
The class interval in which the maximum number of surnames lie is 6 - 8 The weighted frequency of this class interval (taking 2 as the minimum class size) is 44.
Since the class sizes vary to make the histogram we have to calculate the weighted frequency for each rectangle as per its width Minimum class size = 6 - 4 = 2 The modified table showing the weighted frequency as per the size of the class intervals is as follows. The histogram representing the information given in the above table is as follows.
Since the class sizes vary to make the histogram we have to calculate the weighted frequency for each rectangle as per its width Minimum class size = 2 - 1 = 1 The modified table showing the weighted frequency as per the size of the class intervals is as follows. The histogram representing the information given in the above table is as follows.
To make the frequency polygon we first modify the table as follows To make the frequency polygon we mark the marks on the x-axis and the number of students on the y-axis. The representation of the given information in the form of frequency polygon is as follows. From the frequency polygon we can see that the performance of section A is better.
Lamps having life time in the range 700 - 800 = 74 Lamps having life time in the range 800 - 900 = 62 Lamps having life time in the range 900 - 1000 = 48 Lamps having a life time of more than 700 hours = 74 + 62 + 48 = 184.
The representation of the given information in the form of a histogram is as follows.
No it is certainly not correct to conclude that the maximum number of leaves are 153 mm long because the given data does not tell us about the exact length of the leaves. It only tells us about the range in which their lengths lie. We can only conclude that the maximum number of leaves (12) have their lengths in the region 145 - 153.
A frequency polygon could be another suitable graphical representation for the same data.
As we can see from the given table that the data is discontinous and the difference between the upper limit of a class and the lower limit of the next class is 1 and therefore we change both of them by a value 1/2. e.g 127 - 135 would become 126.5 - 235.5 The modified table therefore is The representation of the above data through a histogram is as follows
Party A has won the maximum number of seats. Party A has won 75 seats.
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