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The range of the following ungrouped data will be:
30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 86, 41, 14, 15, 35, 112.
Range = uppermost value – lowest value
= 112 – 14
= 98
The marks of 40 students in final exam obtained by students of class 9 is given below:
8, 18, 12, 6, 8, 16, 12, 5, 23, 2, 16, 23, 2, 10, 20, 12, 9, 7, 6, 5, 3, 5, 13, 21, 13, 15, 20, 24, 1, 7, 21, 16, 13, 18, 23, 7, 3, 18, 17, 16.
Q. The number of students in the class interval 5 – 10 are:
5 – 10
The class marks distribution are:
25, 26, 27, 31, 36, 41, 46, 51, 57, 59
Q. The lower limit of first class interval will be:
Lower limit =
Let ‘l’ be the lower limit of a class interval in a frequency distribution and ‘m’ be the midpoint of the class. Then the upper limit of the class is:
Given that, the lower class limit of a class interval is l and the midpoint of the class is m. Let u be the upper class limit of the class interval. Therefore, we have
⇒ l +u = 2m
⇒ u = 2m  1
Thus the upper class limit of the class is (2m  l).
The mid value of a class interval is 14 and the class size is 2. The lower limit of the class is:
Lower limit = midvalue –
A frequency polygon is constructed by plotting frequency of the class interval and the
A frequency polygon is constructed by plotting frequency of the class interval and mid value of the class.
Cumulative histograms, also known as ogives, are graphs that can be used to determine how many data values lie above or below a particular value in a data set. An ogive (ohjive), sometimes called a cumulative frequency polygon, is a type of frequency polygon that shows cumulative frequencies.
Total number of students
= 4 + 10 + 16 + 20 + 16
= 66
If the mean of x_{1}, x_{2}, x_{3} ……., x_{n} is and 5 is added to each number, the new mean will be:
Given
⇒ if 5 is added to every number, then,
The mean of 10 numbers is 16. If two consecutive numbers are excluded, the new mean is 18. The sum of the excluded numbers is:
⇒ x_{1} + x_{2} …… + x_{10} = 16 ……. (i)
Let x_{2} and x_{3} are removed, then,
⇒ x_{1} + x_{4} + ….. + x_{10} = 144 …….(ii)
Now, Eq. (i) – (ii)
x_{2} + x_{3} = 160 – 144 = 16
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