1). If the sum of 11 consecutive natural numbers is 2761, then the middle number is:
A). 249 |
B). 250 |
C). 251 |
D). 252 |
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2). The mean weight of 9 items is 15. If one more item is added to the series, the mean 16. The value of 10th item is
A). 35 |
B). 30 |
C). 25 |
D). 20 |
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3). The average of the square of the number 0, 1, 2, 3, 4.........n is:
A). \( \Large \frac{1}{2}n \left(n+1\right) \) |
B). \( \Large \frac{1}{6}n \left(2n+1\right) \) |
C). \( \Large \frac{1}{6} \left(n+1\right) \left(2n+1\right) \) |
D). \( \Large \frac{1}{6}n \left(n+1\right) \) |
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4). If the average of number 148, 146, 144, 142,..... in AP be 125, then the total numbers in the series will be
A). 18 |
B). 24 |
C). 30 |
D). 48 |
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5). The mean of the values of 1, 2, 3, ..., n with respectively frequencies x, 2x, 3x, ..., nx is
A). \( \Large \frac{n}{2} \) |
B). \( \Large \frac{1}{3} \left(2n+1\right) \) |
C). \( \Large \frac{1}{6} \left(2n+1\right) \) |
D). \( \Large \frac{n}{3} \) |
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6). The mean of 12 items is \( \Large \bar{X} \) . If the first term is increased by 1, second by 2 and so on, then the new mean is
A). \( \Large \overline{X}+n \) |
B). \( \Large \overline{X}+\frac{n}{2} \) |
C). \( \Large \overline{X}+\frac{n+1}{2} \) |
D). none of these |
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7). If the mean of n observation \( \Large 1^{2},\ 2^{2},\ 3^{2},........n^{2} \), is \( \Large \frac{46}{11}n \), then n is equal to
A). 11 |
B). 12 |
C). 23 |
D). 22 |
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8). In a moderately skewed distribution the values of mean and median are 5 and 6 respectively. The value of mode in such a situation is approximately equal to:
A). 8 |
B). 11 |
C). 16 |
D). none of these |
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9). The value of mean, median and mode coincides, then the distribution is
A). Positive skewness |
B). Symmetrical distribution |
C). Negative skewness |
D). All of these |
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10). The algebraic sum of the deviation of 20 observations measured from 30 is 2. Then mean of observations is
A). 28.5 |
B). 30.1 |
C). 30.5 |
D). 29.6 |
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