Answer to Question #198061 in Statistics and Probability for shahzad

Question #198061

Q) Consider the following ungrouped data:

41   46    7   46   32    5   14   28   48   49    8   49   48   25   41    8   22   46   40   48

Find the following:

a)    Arithmetic mean

b)   Geometric mean

c)    Harmonic mean

d)   Median

e)    Mode

f)     Range

g)    Mean deviation

h)   Variance

i)      Standard Deviation


1
Expert's answer
2021-05-26T09:31:39-0400

a) Arithmetic mean

=41+46+...40+4820=65120=32.55= \frac{41+46+...40+48}{20}= \frac{651}{20}=32.55

b) Geometric mean

=(41×46××40×48)120=26.3908= (41\times 46 \times … \times 40 \times 48)^{\frac{1}{20}} = 26.3908

c) Harmonic mean

=20(141+146+...+140+148)=201.059=18.8854= \frac{20}{ (\frac{1}{41} + \frac{1}{46}+...+ \frac{1}{40}+ \frac{1}{48}) } \\ = \frac{20}{1.059} = 18.8854

d) Median

5 7 8 8 14 22 25 28 32 40 41 41 46 46 46 48 48 48 49 49

Median =40+412=40.5= \frac{40+41}{2} = 40.5

e) Mode = 46 and 48

f) Range = 49-5 = 44

g) Mean deviation

=4132.55+4632.55+...+4032.55+4832.5520=14.395= \frac{|41-32.55|+|46-32.55|+...+|40-32.55|+|48-32.55|}{20} = 14.395

h) Variance =1201((4132.55)2+(4632.55)2+...+(4032.55)2+(4832.55)2)=270.99= \frac{1}{20-1}( (41-32.55)^2+(46-32.55)^2+...+(40-32.55)^2+(48-32.55)^2 ) = 270.99

i) Standard Deviation =270.99=16.461= \sqrt{270.99} = 16.461


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