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Statistics & Data Interpretation

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Q 1. If the sum of 11 consecutive natural numbers is 2761, then the middle number is:

249

250

251

252

Let the first natural number be x

According to the question

x+x+1x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10=2761

=> 11x + 55 = 2761

x = {tex} \frac{2761 - 55}{11} = 246 {/tex}

Middle number = x + 5 = 246 + 5 = 251

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Q 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

35

30

25

20

Total weight of 9 items = 15 x 9 = 135

and total weigth of 10 items = 16 X 10 = 160

Therefore, weight of 10 items = 160 - 135 = 25

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Q 3. The mean of 12 items is {tex} \bar{X} {/tex} . If the first term is increased by 1, second by 2 and so on, then the new mean is

{tex} \overline{X}+n {/tex}

{tex} \overline{X}+\frac{n}{2} {/tex}

{tex} \overline{X}+\frac{n+1}{2} {/tex}

none of these

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Q 4. If the mean of n observation {tex} 1^{2},\ 2^{2},\ 3^{2},........n^{2} {/tex}, is {tex} \frac{46}{11}n {/tex}, then n is equal to

11

12

23

22

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Q 5. The value of mean, median and mode coincides, then the distribution is

Positive skewness

Symmetrical distribution

Negative skewness

All of these

If mean, median and mode coincides, then there is a symmetrical distribution.

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Q 6. For a series the value of mean deviation is 15, the most likely value of its quartile deviation is

12.5

11.6

13

9.7

Since,

{tex} MD = \frac{4}{5}\sigma,\ QD=\frac{2}{3}\sigma {/tex}

{tex} \therefore\ \frac{MD}{QD} = \frac{6}{5} {/tex}

=> {tex} QD = \frac{5}{6} \left(MD\right)=\frac{5}{6} \left(15\right)=12.5 {/tex}

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Q 7. If {tex} \overline{x} {/tex} is the arithmetic mean of n independent variates {tex} x_{1},\ x_{2},\ x_{3}.....,\ x_{n} {/tex} each of the standard deviation {tex} \sigma {/tex}, the variance {tex} \overline{x} {/tex} is

{tex} \frac{\sigma^{2}}{n} {/tex}

{tex} \frac{n\sigma^{2}}{2} {/tex}

{tex} \frac{\left(n+1\right)\sigma^{2}}{3} {/tex}

None of these

We have, {tex} \overline{x} = \frac{1}{n}\sum^{n}_{i=1}x_{i} {/tex}

Therefore, {tex} var \left(\overline{x}\right) = \frac{1}{n^{2}}\left[ \sum^{n}_{i=1} var \left(x_{j}\right)+2\sum^{n}_{i \neq j} cov \left(x_{i},\ x_{j}\right) \right] {/tex}

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Q 8. Calculate the mean for the following frequency distribution.

x: | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

f: | 6 | 8 | 9 | 10 | 7 | 5 | 15 |

54.32

62.24

48.16

42.14

x | f | m | {tex} x' = \frac{m-A}{i} {/tex} | fx' |

10-20 | 6 | 15 | -3 | -18 |

20-30 | 8 | 25 | -2 | -16 |

30-40 | 9 | 35 | -1 | -9 |

40-50 | 10 | 45 | 0 | 0 |

50-60 | 7 | 55 | 1 | 7 |

60-70 | 5 | 65 | 2 | 10 |

70-80 | 15 | 75 | 3 | 45 |

{tex} \sum =60 {/tex} | {tex} \sum fx'=19 {/tex} |

m - middle point of the interval.

i. -differene of the class interval.

A = 45

Arithmetic mean {tex} \overline{x} = A+ \frac{\sum f'x}{\sum f}i {/tex}

= {tex} 45+\frac{19}{60} \times 10 {/tex}

={tex} 45+3.16 = 48.16 {/tex}

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