10th Grade > Mathematics
STATISTICS MCQs
:
B
Class mark is the midpoint of a class interval.
Therefore, its formula is given by upper limit + lower limit2.
:
C
xififixi1022013565167112196114∑fi=20 ∑fixi=311
Mean =∑fixi∑fi
=31120
=15.55
:
xififixi1010100p1515p187126219189259225
∑fi=50
∑fixi = 640+15p
Given mean=17
17 = 640+15p50
Solving, we get p = 14
:
D
Class FrequencyCumulativeFrequency0 - 20101020 - 4071740 - 60163360 - 80p33+p80 -100841+p
Since median is 50, median class is 40-60.
Using
Median=l+[n2−cff]×h,
Here,n=41+p,
⟹ n2=(41+p)2
Now, 50=40+[(41+p)2−1716×20]
⟹10=[(41+p−34)216×20]
⟹10=[(7+p)16×2×20]
Simplifying, we get (7+p)16=1
7+p=16
p=9
:
A
Mode is the most frequently observed value
:
A
The formula for the mode is as follows.
Mode=l+(f1−f02f1−f0−f2)×h
l= lower boundary of the modal class
h= size of the modal class interval
f1= frequency of the modal class.
f0= frequency of the class preceding the modal class
f2= frequency of the class succeeding the modal class
If the preceding and succeeding classes have the same frequency, then f0=f2=f(say).
Then the equation reduces to
Mode=l+(f1−f2f1−f−f)×h
Mode=l+(f1−f2(f1−f))×h
Mode=l+12×h, which is the midpoint of the modal class.
∴ If the preceding and succeeding classes of the modal class have the same frequency, then the mode will be at the midpoint of the modal class.
:
A
In the formula for mode, f1 represents the frequency of the modal class.
:
B
The class mark is the midpoint of the lower and upper limits of a class. It is calculated by taking the average of the upper and lower limits i.e.
Class mark = Upper limit+Lower limit2
:
B
We know Mean = ∑fx∑f
Hence, 5x+212 = 6
⇒ 5x + 2 = 12 × 6
⇒ 5x = 72 – 2 = 70
⇒ x = 14
:
A
We know that,
3 Median = Mode + 2 Mean
or
Mode = 3 Median - 2 Mean
= 3×6−2×7
= 18 - 14
= 4
Therefore, mode = 4