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The mean of 5 observation is 5 and their variance is 12.4. If three of the observations are 1,2 and 6; then the mean deviation from the mean of the data is:

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If the mean deviation 1, 1+d, 1+2d, … , 1+100d from their mean is 255, then d is equal to

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If and , then a possible value of n is among the following is

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The 10th and 50th percentiles of the observation 32, 49, 23, 29, 118 respectively are

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The first three moments of a distribution about 2 are 1, 16, -40 respectively. The mean and variance of the distribution are

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If ${{a}}_1,{{a}}_2,\ldots,{{a}}_n$ are any real numbers and $n$ is
any positive integer, then

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The mean of 25 observations was found to be 38. It was later discovered that 23 and 38 were misread as 25 and 36, then the mean is

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If the mean of the squares of first n natural numbers be 11, then n is equal to?

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The standard deviation of 20 numbers is 30. If each of the numbers is increased by 4, then the new standard deviation will be

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If a variable takes values 0, 1, 2,…, 50 with frequencies $1,\, {{50}}_{{{C}}_1},{{50}}_{{{C}}_2},\ldots..,{{50}}_{{{C}}_{50}}$, then the AM is

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Consider the following frequency distribution table.

Class Interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

Frequency | 180 | $f_1$ | 34 | 180 | 136 | $f_2$ | 50 |

If the total frequency is 686 and the median is 42.6, then the value of $f_1$;and $f_2$ are

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A, B, C are three sets of values of x:

A: 2,3,7,1,3,2,3

B: 7,5,9,12,5,3,8

C: 4,4,11,7,2,3,4

Select the correct statement among the following:

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Increasing Order : A: 1, 2, 2, 3, 3, 3, 7

Mode = 3 (occurs maximum number of times)

Median = 3 (the middle term)

Mean =$\frac{(1+2+2+3+3+3+7)}{7}$

$=\frac{21}{7} = 3$

Hence. Mean=Median=Mode

Standard deviation for the following distribution is

Size of item | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

Frequency | 3 | 6 | 9 | 13 | 8 | 5 | 4 |

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Total number of items in the distribution = Σ f_{i} = 3 + 6 + 9 + 13 + 8 + 5 + 4 = 48.

The Mean (x̅) of the given set = \(\rm \dfrac{\sum f_i x_i}{\sum f_i}\).

⇒ x̅ = \(\rm \dfrac{6\times3+7\times6+8\times9+9\times13+10\times8+11\times5+12\times4}{48}=\dfrac{432}{48}\) = 9.

Let's calculate the variance using the formula: \(\rm \sigma^2 =\dfrac{\sum x_i^2}{n}-\bar x^2\).

\(\rm \dfrac{\sum {x_i}^2}{n}=\dfrac{6^2\times3+7^2\times6+8^2\times9+9^2\times13+10^2\times8+11^2\times5+12^2\times4}{48}=\dfrac{4012}{48}\) = 83.58.

∴ σ^{2} = 83.58 - 9^{2} = 83.58 - 81 = 2.58.

And, Standard Deviation (σ) = \(\rm \sqrt{\sigma^2}=\sqrt{Variance}=\sqrt{2.58}\) ≈ **1.607**.

The mean of 5 observation is 5 and their variance is 124. If three of the observations are 1,2 and 6; then the mean deviation from the mean of the data is:

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For a group of 100 candidates, the mean and standard deviation of scores were found to be 40 and 15
respectively. Later on, it was found that the scores 25 and 35 were misread as 52 and 53 respectively. Then the
corrected mean and standard deviation corresponding to the corrected figures are

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Consider the following frequency distribution table.

Class interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

Frequency | 180 | $f_1$ | 34 | 180 | 136 | $f_2$ | 50 |

If the total frequency is 685 & median is 42.6 then the values of $f_1$ and $f_2$ are

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