# CBSE Class 11th Math 15 – Statistics MCQs

#### Every observation of the n observations in a study got reduced by 5 after checking. Then, the variance

Every observation of the n observations in a study got reduced by 5 after checking. Then, the variance

#### The mean of 40 observations is 44 and variance is 121. Two more observations are added and still the mean and variance are the same. Then, the two new observations added are

The observations added should be (mean + S.D) and (mean – S.D) 44 + 11 & 44 – 11 55 and 33

#### Mean deviation, Quartile deviation, Standard deviation are the commonly used ____.

Measures of central tendencies are single representatives of the given data. They do not talk about the scatter of the data i.e. how each observation is deviated from the central value. The scattering is called the dispersion.

#### The curves that are used to determine median are called ____.

There are two types of ogives called, less than ogive and greater than ogive. The point of intersection of both the ogives gives the median of the distribution.

#### The coefficient of range helps us to understand the measure of variability in a frequency distribution. The coefficient of range is defined as

Formula Application

#### The measure of central tendency that is not commonly used to determine the measure of dispersion is ____.

Mode is not based on all observations and further mathematical treatment is not possible.

#### The less than and the more than ogives of a frequency distribution intersect at (42.1, 900). The coefficient of mean deviation about median is 0.643, then mean deviation about median is

Median is the x-coordinate of the point of intersection. Coefficient of mean deviation about median = mean deviation about median/ median 0.643 = x / 42.1 0.643 (42.1) = x x = 27.07

#### The S.D of a frequency distribution is 3.92. The variance of the distribution (Correct to 2 decimal places) is

Variance = Square of S.D = (3. 92)^{2}=15.37 (Approximately)

#### For an unskewed frequency distribution the empirical formula connecting the three measures of central tendancy is

Emperical Formula Mode = 3 Median – 2 Mean

#### In a discreet frequency distribution, if each observation is increased or decreased by a constant value, then the measures of dispersion, mean deviation and standard deviation is ____.

When each observation is increased or decreased by a constant value, then mean or median also gets increased or decreased by the same amount. Therefore, the deviation of each observation remains the same.

#### The sum of the numerical values of mean deviation about median is

Property of Median

#### Algebraic sum of mean deviations is always ____

the mean of 5 observations = x_{1} + x_{2} + x_{3} + x_{4 }+ x_{5}/5 = x ⇒ x_{1 }+ x_{2} + x_{3} + x_{4 }+ x_{5} = 5x ⇒ x_{1 }+ x_{2} + x_{3} + x_{4 }+ x_{5} - x) = 0

#### The mean and variance of 4, 5, 6, 7...20 are

Sum of n natural numbers = n(n + 1)/2 4 + 5 + 6 + .... + 20 = Sum upto 20 - Sum upto 3 = ((20 × 21)/ 2) - (1 + 2 + 3) = 210 - 6 = 204 Mean = 204/ 7 = 12 4^{2} + 5^{2} + 6^{2} + .... + 20^{2} = Σ20^{2} - Σ3^{2} = (1/6) (20) (21) 41 – 14 = 2870 - 14 =2856 Variance =( ∑x^{2}/n) -x^{2} = (2856/17) - 12^{2 }= 168 - 144 = 24

#### The range of statistical data can also be used to measure variability. The coefficient of range is defined as ____.

L is the largest value of the observation and S is the smallest value of the observation. The coefficient of range = L – S/L + S

#### The coefficient of variance for a frequency distribution is 45.55%. The mean of the distribution is 20. Then, the variance is

C.V =(S.D/Mean) × 100 45.55 = (S.D/20) × 100 S.D = (45.55)20/100 = 9.11 VARIANCE = (S. D)^{2} = (9.11)^{2 }= 82.9921

#### In a mid-term test, a child scored 12, 14, 20, 8, 23 in Maths, Science, English, Hindi, Social studies respectively. The scores are for 25 in each subject but for report card entry, the scores are to be transformed to 50. Then, the new variance is

Mean = (12 + 14 + 20 + 8 + 23) /5 = 15.4 Variance =[(∑x^{2}) /n ]-x^{2} (144 +196 +400 +64 + 529 )/5 – (15.4)^{2}= 266.6 – 237.16 = 29.44 New marks are 24, 28, 40, 16, 46 New mean = 154/5 = 30.8 New variance = 5332/5 – (30. 8)^{2 } = 1066.4 – 948.64 = 117 .76 = 4( 29.44) Hence, the new variance is 4 times the original.

#### The mean of 5 observations is 5 and the variance is 6.8. Three of the observations are 2, 3, 9. Then, the other 2 observations are

Mean = ( 2 + 3 + 9 + x + y)/5 = 5 x + y = 25 - 14 = 11 Variance = ((4 + 9 + 81 + x^{2} + y^{2})/5) -25 = 6.8 x^{2} + y^{2}=( 31.8) × 5 – 94 x^{2} + y^{2}= 65 x + y = 11 x^{2} + y^{2}+ 2xy = 121 2xy = 121 - 65 = 56 xy = 28 x = 28/y 28/y +y =11 28+ y^{2} =11y y^{2} - 11y + 28 = 0 (y-4)(y-7) = 0 y = 4 Therefore, the two values are 4 and 7.

#### The variance is ____.

The variance of a variate x is the arithmetic mean of the squares of all deviations of x from the arithmetic mean of the observations.

#### The most powerful measure of dispersion is ____.

In determining the mean deviation about mean and mean deviation about median, we use absolute values of the deviation. In determining S.D, the deviations are squared and then the square root is taken. Hence, it can be used for further analysis. The unit is squared in variance. Therefore, S.D is the most powerful.

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