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The no. of class intervals, if the magnitude of class interval is 4 will be:
Data: 31, 23, 19, 29, 20, 16, 22, 10, 13, 34, 33, 38, 36, 24, 18, 15, 12, 30, 27, 23, 20
Range = 38 – 10 = 28
∴ number of class intervals = 28/4 = 7
The class marks of a distribution are:
52, 47, 57, 67, 62, 72, 82, 87, 97, 92, 102.
Q. The lower and upper limits of first class interval will be:
Upper limit =
Lower limit =
= 44.5
[h = difference between any two marks = 52 – 47 = 5]
Tallies are usually marked in groups of five. The first four tallies are marked vertically and the fifth tally is marked diagonally across the first 4.
The mid–value and upper limit of a class interval are 41 and 47 respectively. The class size will be:
Upper limit = mid value +
47 = 41 +
∴ Class size = (47 – 41) × 2
= 12
xaxis represents class interval, yaxis represents frequency.
In the ‘more–than’ type of ogive the cumulative frequency is plotted against:
More than distribution of cumulative frequency plot is a kind of Ogive frequency distribution in which the frequencies are decreasing from top to bottom.
So, we consider lower limits as abscissa and more than cumulative frequencies as ordinate.
The frequency of students is highest in the class interval:
∵ The class interval, 30 – 40, has highest peak,
∴ It has highest frequency.
The mean of x_{1}, x_{2}, ……., x_{n} is , then the value of:
⇒ x_{1} + x_{2} + x_{3} + x_{4} ……. + x_{n} = n
⇒ x_{1} + x_{2} + x_{3} …… + x_{n}
If each number in (Prob  17) is multiplied by k, the new mean will be:
Given
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
The sum of deviations of a set of n values x_{1}, x_{2}, …… x_{n} measured from 50 is – 10 and the sum of deviations of the values from 46 is 70. The value of n is:
…(i)
…(ii)
Subtracting eq. (i), from eq. (ii),
46n – (–50 n) = 70 – (–10)
⇒ 4n = 80
⇒ n = 20
f_{1} + f_{2} + 17 + 32 + 19 = 120
⇒ f_{1} + f_{2} = 52 …(i)
Mean
⇒ 50 × 120 = 170 + 1600 + 1710 + + 30f_{1} + 70f_{2}
⇒ 30f_{1} + 70f_{2} = 2520
⇒ 37_{1} + 7f_{2} = 252 ……..(ii)
From eq (i) and eq (ii).
f_{1} = 28, f_{2} = 24
The new median, of the following data, if 37 is replaced by 5.
7, 9, 16, 25, 31, 36, 37, 39, 40, 42, 43
If 37 ⟶ 5, then, the new data will be:
5, 7, 9, 16, 25, 31, 36, 39, 40, 42, 43
∵ No. of numbers = 11
∴ = 6^{th} number will be the median.
The mode of the following data is:
29, 40, 41, 46, 45, 44, 43, 29, 40, 41, 46, 44, 44, 47, 49, 53, 29, 57, 44, 43, 41, 28, 16, 26.
∵ 44 is repeated 4 times, i.e., maximum no. of times.
∴ Mode = 44
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