12th Grade > Mathematics
STATISTICS AND COLLECTION OF STATISTICAL DATA MCQs
Statistics, Introduction To Statistics, Collection Of Data (11th And 12th Grade)
Total Questions : 105
| Page 8 of 11 pages
Answer: Option C. -> 57 kg
:
C
Total weight of 7 students is =55×7=385kg
Sum of weight of 6 students
=52+58+55+53+56+54=328kg
∴Weight of seventh student =385–328=57kg.
:
C
Total weight of 7 students is =55×7=385kg
Sum of weight of 6 students
=52+58+55+53+56+54=328kg
∴Weight of seventh student =385–328=57kg.
Answer: Option A. -> 100×s.dmean
:
A
C.V.isdefinedbyC.V=100×s.dmean
:
A
C.V.isdefinedbyC.V=100×s.dmean
Answer: Option C. -> 1
:
C
Population B observations are 151, 152, ….., 250
Let y, be these observations
Shifting the origin zi=yi−50
The values of these observations are 100, 101, ….., 200.
∴VA=VB
( ∵ Variance is independent of shifting the origin).
:
C
Population B observations are 151, 152, ….., 250
Let y, be these observations
Shifting the origin zi=yi−50
The values of these observations are 100, 101, ….., 200.
∴VA=VB
( ∵ Variance is independent of shifting the origin).
Answer: Option A. -> 75
:
A
Given,
82=(27+x)+(31+x)+(89+x)+(107+x)+(156+x)5
⇒82×5=410+5x⇒410−410=5x⇒x=0
∴Required mean is,
¯¯¯x=130+x+126+x+68+x+50+x+1+x5
¯¯¯x=375+5x5=375+05=3755=75.
:
A
Given,
82=(27+x)+(31+x)+(89+x)+(107+x)+(156+x)5
⇒82×5=410+5x⇒410−410=5x⇒x=0
∴Required mean is,
¯¯¯x=130+x+126+x+68+x+50+x+1+x5
¯¯¯x=375+5x5=375+05=3755=75.
Answer: Option B. -> 39.95, 14.98
:
B
Corrected∑x=40×200−50+40=7990
∴Corrected¯¯¯x=7990200=39.95
∑x2=n[σ2+¯¯¯x2]=200[152+402]=365000
Correct∑x2=365000−2500+1600=364100
∴Correctedσ=√364100200−(39.95)2
=√(1820.5−1596)=√224.5=14.98.
:
B
Corrected∑x=40×200−50+40=7990
∴Corrected¯¯¯x=7990200=39.95
∑x2=n[σ2+¯¯¯x2]=200[152+402]=365000
Correct∑x2=365000−2500+1600=364100
∴Correctedσ=√364100200−(39.95)2
=√(1820.5−1596)=√224.5=14.98.
Answer: Option C. -> ¯¯¯x+10αα
:
C
Let x1,x2……,xn be n observations.
Then¯¯¯x=1n∑xi
letyi=xiα+10
¯¯¯y=1nn∑iyi
And,¯¯¯y=1n[(n∑ixiα+10.n]
⇒¯¯¯y=1α¯¯¯x+10
=¯¯¯x+10αα.
:
C
Let x1,x2……,xn be n observations.
Then¯¯¯x=1n∑xi
letyi=xiα+10
¯¯¯y=1nn∑iyi
And,¯¯¯y=1n[(n∑ixiα+10.n]
⇒¯¯¯y=1α¯¯¯x+10
=¯¯¯x+10αα.
Answer: Option A. -> α−54
:
A
Arrange the data as: α−72,α−3,α−52,α−2,α−12,α+12,α+4,α+5,
Median =a−2+α−122=2α−522=α−54
:
A
Arrange the data as: α−72,α−3,α−52,α−2,α−12,α+12,α+4,α+5,
Median =a−2+α−122=2α−522=α−54
Answer: Option C. -> √3
:
C
Given that median = 4 and mode = 6. We know that
Mean - mode = 3 (Mean - Median)
Mean - 6 = 3 (Mean - 4)
¯x−6=3¯x−12
¯x=3
σ2=∑x2in−(∑xin)2=484−(3)2=3 ∴σ=√3
:
C
Given that median = 4 and mode = 6. We know that
Mean - mode = 3 (Mean - Median)
Mean - 6 = 3 (Mean - 4)
¯x−6=3¯x−12
¯x=3
σ2=∑x2in−(∑xin)2=484−(3)2=3 ∴σ=√3
Answer: Option D. -> 31
:
D
20∑i=1(xi−30)=20=20∑i=1xi−20×30=20
⇒20∑i=1xi=620.Mean=20∑i=120=62020=31.
:
D
20∑i=1(xi−30)=20=20∑i=1xi−20×30=20
⇒20∑i=1xi=620.Mean=20∑i=120=62020=31.
Answer: Option B. -> ¯x+n+1
:
B
¯x=∑ni=1xin⇒∑ni=1xi=n¯x∴∑ni(xi+2i)n=∑ni=1xi+2(1+2+...+n)n=n¯x+2n(n+1)2n=¯x+(n+1)
:
B
¯x=∑ni=1xin⇒∑ni=1xi=n¯x∴∑ni(xi+2i)n=∑ni=1xi+2(1+2+...+n)n=n¯x+2n(n+1)2n=¯x+(n+1)
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