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Measures of Dispersion Test - 3

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Measures of Dispersion Test - 3
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  • Question 1
    1 / -0

    Absolute measures of dispersion are

    Solution

    These measures give us an idea about the amount of dispersion in a set of observations. They give the answers in the same units as the units of the original observations. When the observations are in kilograms, the absolute measure is also in kilograms.

  • Question 2
    1 / -0

    The main disadvantage of the standard deviation is

    Solution

    It is very hard for a person to calculate square root and even square of deviations. Formula of standard deviation is very hard to remember.

  • Question 3
    1 / -0

    The variance

    Solution

    Since variance is a relative measures which is free from the units in which the values have been expressed, they can be compared even across different groups having different units of measurement.

  • Question 4
    1 / -0

    The inter quartile range

    Solution

    In descriptive statistics, the interquartile range (IQR), also called the midspread or middle 50%, or technically H-spread, is a measure of statistical dispersion, being equal to the difference between 75th and 25th percentiles, or between upper and lower quartile. IQR = Q3 −  Q1

  • Question 5
    1 / -0

    A measure of relative dispersion is given by the: 

    Solution

    Relative measures of dispersion, are also known as coefficients of dispersion, are obtained as ratios or percentages. The coefficient of variation (CV) is the ratio of the standard deviation to the mean (average).

  • Question 6
    1 / -0

    The heights in cm of a group of first year biology students were recorded. The variance of these heights was subsequently calculated. The unit of measurement for this variance is:

    Solution

    As the variance is mean of sum of square deviation so unit will be in squared cm.

  • Question 7
    1 / -0

    The number of protozoa seen under 5 separate microscope slides were 3,4,5,6 and 7. The sum of squared deviations of observations from their sample mean is:

    Solution

    sum of squared deviations of observations = ∑(xi-μ)2

    μ=  ∑Xi/N = (3+4+5+6+7)/5= 25/5= 5

    (xi-μ)= -2, -1, 0, 1, 2

     ∑(xi-μ)2= 4+1+0+1+4= 10

  • Question 8
    1 / -0

    What would happen to the variance of a data set if we added 5 to every observation?

    Solution

    variance is unaffected due to change of origin.

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