Q1: If the median of the distribution is 28.5, find thevalues of x1 and x2.
(a) 8,7
(b) 9,7
(c) 6,8
(d) 7,9
Ans: (a)
Where L is lower limit of class 20−30
f is frequency of class 20−30,
w is width of class
C is cumulative frequency of previous class andN = ΣF
x1 = 8
From eq. (i), we get
x1 + x2 =15
8 + x2 = 15 x 2 = 7
∴ value of x1 & x2 are 8 and 7 respectively.
Q2: The table shows the distribution of the scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution.
Use your graph to estimate the median.
Ans:
Since, no. of terms, n=160 which is even
From the graph, me = 43.5
Q3: The median of the distribution given is 46. Find the values of x and y, if the total frequency is 230.
Ans:
⇒ 73 − x = 39
⇒ x = 34 ∴ y = 46
Q4: Six numbers from a list of nine integers are 7,8,3,5,9 and 5. Find the largest possible value of the median of all nine numbers in this list.
Ans:
⇒ The given six numbers are 7,8,3,5,9 and 5.
⇒ Arrange the six numbers in order 3,5,5,7,8,9.
⇒ We have three more numbers to insert into the list, and the median will be the highest 5(and 5 lowest) number on the list. If the three numbers are greater than 9, the median will be the highest it can possible be.
∴ The median is the 5th item of data which is 8.
∴ Median = 8
Q5: The mean of the numbers 1,2,3,...... .,n with respective weights 12 + 1, 22 +2,....n2 + n is
(a)
(b)
(c)
(d)
Ans: (c)
Given xi = i and weights wi = i2 + i(i = 1, 2, 3,...n)
Q6: A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Ans: To find the class mark (xi) for each interval, using the following relation:
Taking 17 as assumed mean (a), calculating di and fidi as:
From the table:
= 17 − 4.525
= 12.475
≈ 12.48
Therefore, the mean number of days = 12.48 days for which a student was absent.
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