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Quiz 1
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The sum of 10 items is 12 and the sum of their squares isThe standard deviation is
0%
1/5
0%
2/5
0%
3/5
0%
4/5
Explanation
3/5
Hint:
Given, ∑x = 12 and ∑x² = 18
Now, varience = ∑x²/n – (∑x/n)²
⇒ varience = 18/10 – (12/10)²
⇒ varience = 9/5 – (6/5)²
⇒ varience = 9/5 – 36/25
⇒ varience = (9 × 5 – 36)/25
⇒ varience = (45 – 36)/25
⇒ varience = 9/25
⇒ Standard deviation = √(9/25)
⇒ Standard deviation = 3/5
The algebraic sum of the deviation of 20 observations measured from 30 isSo, the mean of observations is
0%
30.0
0%
30.1
0%
30.2
0%
30.3
Explanation
30.1
Hint:
Given, algebraic sum of of the deviation of 20 observations measured from 30 is 2
⇒ ∑(x
i
– 30) = 2 {1 ≤ i ≤ 20}
⇒ ∑x
i
– 30 × 20 = 2
⇒ (∑x
i
)/20 – (30 × 20)/20 = 2/20
⇒ (∑x
i
)/20 – 30 = 0.1
⇒ Mean – 30 = 0.1
⇒ Mean = 30 + 0.1
⇒ Mean = 30.1
The coefficient of variation is computed by
0%
S.D/.Mean×100
0%
S.D./Mean
0%
Mean./S.D×100
0%
Mean/S.D.
Explanation
S.D./Mean
Hint:
The coefficient of variation = S.D./Mean
When tested the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494,The mean of the lives of 5 bulbs is
0%
1445
0%
1446
0%
1447
0%
1448
Explanation
1446
Hint:
Given, lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623
Now, mean = (1357 + 1090 + 1666 + 1494 + 1623)/5
= 7230/5
= 1446
If mode of a series exceeds its mean by 12, then mode exceeds the median by
0%
4
0%
8
0%
6
0%
12
Explanation
8
Hint:
Given, Mode = Mean + 12
⇒ Mode – 12 = Mean
Now, Mode = 3×Median – 2×Mean
⇒ Mode = 3×Median – 2(Mode – 12)
⇒ Mode = 3×Median – 2×Mode + 24
⇒ Mode + 2×Mode = 3×Median + 24
⇒ 3×Mode = 3×Median = 24
⇒ Mode = Median + 8
So, mode exceeds the median by 8
The median and SD of a distributed are 20 and 4 respectively. If each item is increased by 2, the new median and SD are
0%
20, 4
0%
22, 6
0%
22, 4
0%
20, 6
Explanation
22, 4
Hint:
Since each value is increased by 2, therefore the median value is also increased by 2. So, new median = 22
Again, the variance is independent of the change of origin. So it remains the same.
Range of the data 4, 7, 8, 9, 10, 12, 13 and 17 is
0%
4
0%
17
0%
13
0%
21
Explanation
13
Hint:
Give, data are: 4, 7, 8, 9, 10, 12, 13 and 17
Range = Maximum value – Minimum Value
= 17 – 4
= 13
If Mean = Median = Mode, then it is
0%
Symmetric distribution
0%
Asymmetric distribution
0%
Both symmetric and asymmetric distribution
0%
None of these
Explanation
Symmetric distribution
Hint:
In a symmetric distribution,
Mean = Median = Mode
If the difference of mode and median of a data is 24, then the difference of median and mean is
0%
12
0%
24
0%
8
0%
36
Explanation
12
Hint:
Given the difference of mode and median of a data is 24
⇒ Mode – Median = 24
⇒ Mode = Median + 24
Now, Mode = 3×Median – 2×Mean
⇒ Median + 24 = 3×Median – 2×Mean
⇒ 24 = 3×Median – 2×Mean – Median
⇒ 24 = 2×Median – 2×Mean
⇒ Median – Mean = 24/2
⇒ Median – Mean = 12
If r is the correlation coefficient, then
0%
|r| ≤ 1
0%
r ≤ 1
0%
|r| ≥ 1
0%
r ≥ 1
Explanation
|r| ≤ 1
Hint:
If r is the correlation coefficient, then |r| ≤ 1
If the mean of the following data is 20.6, then the value of p is
x = 10 15 p 25 35
f = 3 10 25 7 5
0%
30
0%
20
0%
25
0%
10
Explanation
20
Hint:
Mean = ∑ f
i
× x
i
/∑ f
i
⇒ 20.6 = (10 × 3 + 15 × 10 + p × 25 + 25 × 7 + 35 × 5)/(3 + 10 + 25 + 7 + 5)
⇒ 20.6 = (30 + 150 + 25p + 175 + 175)/50
⇒ 20.6 = (530 + 25p)/50
⇒ 530 + 25p = 20.6 × 50
⇒ 530 + 25p = 1030
⇒ 25p = 1030 – 530
⇒ 25p = 500
⇒ p = 500/25
⇒ p = 20
So, the value of p is 20
If the mean of first n natural numbers is 5n/9, then n =
0%
5
0%
4
0%
9
0%
10
Explanation
9
Hint:
Given mean of first n natural number is 5n/9
⇒ (n+1)/2 = 5n/9
⇒ n + 1 = (5n×2)/9
⇒ n + 1 = 10n/9
⇒ 9(n + 1) = 10n
⇒ 9n + 9 = 10n
⇒ 10n – 9n = 9
⇒ n = 9
Which one is measure of dispersion method
0%
Renge
0%
Quartile deviation
0%
Mean deviation
0%
all of the above
Explanation
all of the above
Hint:
Range, Quartile deviation, Mean deviation all are the measure of dispersions method.
If a variable takes discrete values x + 4, x – 7/2, x – 5/2, x – 3, x – 2, x + 1/2, x – 1/2, x + 5 (x > 0), then the median is
0%
x – 5/4
0%
x – 1/2
0%
x – 2
0%
x + 5/4
Explanation
x – 5/4
Hint:
Given, discrete values x + 4, x – 7/2, x – 5/2, x – 3, x – 2, x + 1/2, x – 1/2, x + 5
Now, arrange them in ascending order, we get
x – 7/2, x – 3, x – 5/2, x – 2, x – 1/2, x + 1/2, x + 4, x + 5
Total observations = 8
Now, median = AM of 4th and 5th observations
= AM of (x – 2) and (x – 1/2) observations
= (x – 2 + x – 1/2)/2
= (2x – 5/2)/2
= x – 5/4
If covariance between two variables is 0, then the correlation coefficient between them is
0%
nothing can be said
0%
0
0%
positive
0%
negative
Explanation
0
Hint:
The relationship between the correlation coefficient and covariance for two variables as shown below:
r
(x, y)
= COV(x, y)/{s
x
× s
y
}
r
(x, y)
= correlation of the variables x and y
COV(x, y) = covariance of the variables x and y
s
x
= sample standard deviation of the random variable x
s
x
= sample standard deviation of the random variable y
Now given COV(x, y) = 0
Then r
(x, y)
= 0
The mean of a group of 100 observations was found to beLater on, it was found that three observations were incorrect, which was recorded as 21, 21 andThen the mean if the incorrect observations are omitted is
0%
18
0%
20
0%
22
0%
24
Explanation
20
Hint:
Given mean of 100 observations is 20
Now
∑ xi/100 = 20 (1 = i = 100)
⇒ ∑xi = 100×20
⇒ ∑xi = 2000
3 observations 21, 21 and 18 are recorded in-correctly.
So ∑xi = 2000 – 21 – 21 – 18
⇒ ∑xi = 2000 – 60
⇒ ∑xi = 1940
Now new mean is
∑ xi/100 = 1940/97 = 20
So, the new mean is 20
Varience is independent of change of
0%
origin only
0%
scale only
0%
origin and scale both
0%
None of these
Explanation
origin only
Hint:
Varience is independent of change of origin only.
Let x
1
, x
2
, x
3
, ……… , x
n
, be n observations and X be the arithmetic mean. Then formula for the standard deviation is given by
0%
∑(x
i
– mean)²
0%
∑(x
i
– mean)2 /n
0%
√{∑(x
i
– mean)²/n}
0%
None of these
Explanation
√{∑(x
i
– mean)²/n}
Hint:
Given, x
1
, x
2
, x
3
, ………. , x
n
be n observations and X be the arithmetic mean.
Now standard deviation = √{∑(x
i
– mean)²/n}
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