Statistics Practice Questions - DPP for JEE

 Page 2


3. (b) On arranging the given observations in ascending order, we get
All negative terms  All positive terms
The median of given observations = (n+1)
th
term = 0
? S.D.>M.D.
4. (a) First ten positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
Sum of these numbers = 1 + 2 + ... + 10 = 55
Sum of squares of these numbers 
= 1
2
 + 2
2
 + … + 10
2
 = 385
Standard deviation 
= 
? Variance (s
2
) = 8.25
5. (a) We know that,
Coefficient of variation = 
? CV of 1st distribution = 
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3. (b) On arranging the given observations in ascending order, we get
All negative terms  All positive terms
The median of given observations = (n+1)
th
term = 0
? S.D.>M.D.
4. (a) First ten positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
Sum of these numbers = 1 + 2 + ... + 10 = 55
Sum of squares of these numbers 
= 1
2
 + 2
2
 + … + 10
2
 = 385
Standard deviation 
= 
? Variance (s
2
) = 8.25
5. (a) We know that,
Coefficient of variation = 
? CV of 1st distribution = 
?  [CV of 1st distribution = 50 (given)]
? s
1
 = 15
Also, CV of 2nd distribution = 
? ?  ? s
2
 = 15
Thus, s
1
 – s
2
 = 15 – 15 = 0
6. (b) Corrected Sx = 40 × 200 – 50 + 40 = 7990
? Corrected = 7990 / 200 = 39.95
Incorrect Sx
2
 = n[s
2
 + ] = 200[15
2
 + 40
2
]= 365000
Corrected Sx
2 
= 365000 – 2500 + 1600 = 364100
? Corrected 
7. (b) Let 
So,  and hence
 
 
Hence standard deviation is multiplied by 
8. (b) Let the observations be x
1
, x
2
, ...., x
20
 and  be their mean. Given
that, variance = 5 and n = 20. We know that,
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3. (b) On arranging the given observations in ascending order, we get
All negative terms  All positive terms
The median of given observations = (n+1)
th
term = 0
? S.D.>M.D.
4. (a) First ten positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
Sum of these numbers = 1 + 2 + ... + 10 = 55
Sum of squares of these numbers 
= 1
2
 + 2
2
 + … + 10
2
 = 385
Standard deviation 
= 
? Variance (s
2
) = 8.25
5. (a) We know that,
Coefficient of variation = 
? CV of 1st distribution = 
?  [CV of 1st distribution = 50 (given)]
? s
1
 = 15
Also, CV of 2nd distribution = 
? ?  ? s
2
 = 15
Thus, s
1
 – s
2
 = 15 – 15 = 0
6. (b) Corrected Sx = 40 × 200 – 50 + 40 = 7990
? Corrected = 7990 / 200 = 39.95
Incorrect Sx
2
 = n[s
2
 + ] = 200[15
2
 + 40
2
]= 365000
Corrected Sx
2 
= 365000 – 2500 + 1600 = 364100
? Corrected 
7. (b) Let 
So,  and hence
 
 
Hence standard deviation is multiplied by 
8. (b) Let the observations be x
1
, x
2
, ...., x
20
 and  be their mean. Given
that, variance = 5 and n = 20. We know that,
Variance 
i.e.  or      ...(i)
If each observation is multiplied by 2 and the new resulting observations
are y
i
, then
y
i
 = 2x
i 
i.e., x
i
 = y
i
Therefore, 
i.e.,  or 
On substituting the values of x
i
 and  in eq. (i), we get
 i.e., 
Thus, the variance of new observations
= 
9. (b) The mean of the series
= 
Therefore, mean deviation from mean
= 
10. (a) Mean and SD s of the combined group are
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3. (b) On arranging the given observations in ascending order, we get
All negative terms  All positive terms
The median of given observations = (n+1)
th
term = 0
? S.D.>M.D.
4. (a) First ten positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
Sum of these numbers = 1 + 2 + ... + 10 = 55
Sum of squares of these numbers 
= 1
2
 + 2
2
 + … + 10
2
 = 385
Standard deviation 
= 
? Variance (s
2
) = 8.25
5. (a) We know that,
Coefficient of variation = 
? CV of 1st distribution = 
?  [CV of 1st distribution = 50 (given)]
? s
1
 = 15
Also, CV of 2nd distribution = 
? ?  ? s
2
 = 15
Thus, s
1
 – s
2
 = 15 – 15 = 0
6. (b) Corrected Sx = 40 × 200 – 50 + 40 = 7990
? Corrected = 7990 / 200 = 39.95
Incorrect Sx
2
 = n[s
2
 + ] = 200[15
2
 + 40
2
]= 365000
Corrected Sx
2 
= 365000 – 2500 + 1600 = 364100
? Corrected 
7. (b) Let 
So,  and hence
 
 
Hence standard deviation is multiplied by 
8. (b) Let the observations be x
1
, x
2
, ...., x
20
 and  be their mean. Given
that, variance = 5 and n = 20. We know that,
Variance 
i.e.  or      ...(i)
If each observation is multiplied by 2 and the new resulting observations
are y
i
, then
y
i
 = 2x
i 
i.e., x
i
 = y
i
Therefore, 
i.e.,  or 
On substituting the values of x
i
 and  in eq. (i), we get
 i.e., 
Thus, the variance of new observations
= 
9. (b) The mean of the series
= 
Therefore, mean deviation from mean
= 
10. (a) Mean and SD s of the combined group are
Thus, AM is decreased by 
(approx)
Hence, (a) is the correct answer.
11. (c) Using, 
12. (a) Given, 
Let a student obtains  out of 75. Then his marks out of 100 are 
Each observation is multiplied by 
New  Variance = 
13. (a)