Mean \(\bar { x } =\frac { 2+3+5+6+9+10+12+17+20+26 }{ 10 } \)
\(=\frac { 110 }{ 10 } =11\)

\(\therefore \) Mean deviation from mean = \(\frac { \sum { |D| } }{ n } =\frac { 62 }{ 10 } =6.2\)

Let E₁ = missing card is a heart card
E₂ = missing card is a spade card
E₃ = missing card is a club card
E₄ = missing card is a diamond card
and A = drawing 2 heart cards from the remaining cards.
\(\therefore\) P(E₁) = P(E₂) = P(E₃) = P(E₄)
\(=\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \)
P(A/E₁) = P(two heart cards given that one heart card is missing)
\(=\frac { 12{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { \frac { 12\times 11 }{ 2\times 1 } }{ \frac { 51\times 50 }{ 2\times 1 } } =\frac { 66 }{ 1275 } \)
P(A/E₂) = P(2 heart cards given that one spade card is missing)
\(=\frac { 13{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { \frac { 13\times 12 }{ 2 } }{ \frac { 51\times 50 }{ 2\times 1 } } =\frac { 78 }{ 1275 } \)
Similarly, P(A/E₃) = \(\frac { 13{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { 78 }{ 1275 } \)
and P(A/E₄) = \(\frac { 13{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { 78 }{ 1275 } \)
\(\therefore\) By Baye's theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 }).P\left( A/{ E }_{ 1 } \right) }{ P({ E }_{ 1 }).P\left( A/{ E }_{ 1 } \right) +P({ E }_{ 2 }).P\left( A/{ E }_{ 2 } \right) +P({ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 4 } \times \frac { 66 }{ 1275 } }{ \frac { 1 }{ 4 } \times \frac { 66 }{ 1275 } +\frac { 1 }{ 4 } \times \frac { 78 }{ 1275 } +\frac { 1 }{ 4 } \times \frac { 78 }{ 1275 } +\frac { 1 }{ 4 } \times \frac { 78 }{ 1275 } } \)
\(=\frac { \frac { 1 }{ 4 } \times \frac { 66 }{ 1275 } }{ \frac { 1 }{ 4 } \left[ \frac { 66 }{ 1275 } +\frac { 78 }{ 1275 } +\frac { 78 }{ 1275 } +\frac { 78 }{ 1275 } \right] } \)
\(=\frac { 66 }{ 1275 } \times \frac { 1275 }{ 66+78+78+78 } \)
\(=\frac { 66 }{ 300 } =\frac { 11 }{ 50 } \)

Let E₁ = Taking out a fair coin from the bag
E₂ = Taking out a coin having head on both sides
A = Tossed coin results in a head
\(\therefore P({ E }_{ 1 })=\frac { n+1 }{ 2n+1) } \) and P(E₂)
\(=\frac { n }{ 2n+1 } \)
Also, P(A/E₁) = \(\frac { 1 }{ 2 } \) and P(A/E₂) = 1
By total probability theorem,
P(A) = P(E₁).P(A/E₁) +P(E₂).P(A/E₂)
\(=\frac { n+1 }{ 2n+1 } \times \frac { 1 }{ 2 } +\frac { n }{ 2n+1 } \times 1\)
\(=\frac { n+1 }{ 2(2n+1) } +\frac { n }{ 2n+1 } \)
\(=\frac { n+1+2n }{ 2(2n+1) } \)
\(=\frac { 3n+1) }{ 2(2n+1) } \)
But it is given that P(A)=\(\frac { 31 }{ 42 } \)
\(\therefore \frac { 31 }{ 42 } =\frac { 3n+1 }{ 2(2n+1) } \)
\(\Rightarrow\) 31(2n+1) = 21(3n+1)
\(\Rightarrow\) 32n + 31 = 63n + 21
\(\Rightarrow\) 31 - 21 = 63n - 32n
\(\Rightarrow\) 10 = n
\(\therefore\) n = 10

Harmonic Mean = \(\frac { n }{ \sum { \frac { 1 }{ x } } } \)
\(=\frac { 4 }{ 0.3190 } =12.54\)
\(\therefore\) HM = 12.54

\(\sum { log } \) = 4.3167
Geometric Mean = Antilog\(\left( \frac { \sum { log\quad x } }{ N } \right) =Antilog\left( \frac { 4.3167 }{ 4 } \right) \)
= Antilog (1.0791)
GM = 11.99

Let us define the following events.
E₁: X speaks truth
E₂: X tells a lie
E: X reports a six
From the data given in the problem, we have
P(E₁) = \(\frac{4}{5}\); P(E₂) = \(\frac{1}{5}\);
P(E/E₁) = \(\frac{1}{6}\); P(E/E₂) = \(\frac{5}{6}\)
The required probability that actually there was six (by Bayes theorem) is 
\(P(E_1/E)=\frac{P(E_1)P(E/E_2)}{P(E_1)P(E/E_1)+P(E_2)P(E/E_2)}\) = \(\frac{\frac{4}{5}\times \frac{1}{6}}{(\frac{4}{5}\times \frac{1}{6})+(\frac{1}{5}\times \frac{5}{6})}=\frac{4}{9}\)

Marks are arranged in ascending order
12 15 20 28 30 40 50
n = number of observations =7
Q₁ = Size of \({ \left( \frac { \left( n+1 \right) }{ 4 } \right) }^{ th }\)value = Size of \({ \left( \frac { 7+1 }{ 4 } \right) }^{ th }\)value = Size of 2^nd value = 15
Hence Q₁ = 15
Q₃ = Size of \({ \left( \frac {3 \left( n+1 \right) }{ 4 } \right) }^{ th }\) value = Size of \({ \left( \frac { 3\times 8 }{ 4 } \right) }^{ th } \)value = Size of 6^th value = 40
Hence Q₃ = 40
QD =\(\quad \frac { 1 }{ 2 } \left( { Q }_{ 3 }-{ Q }_{ 1 } \right) \frac { 40-15 }{ 2 } =12.5\)
Coefficient of QD =\(\frac { { Q }_{ 3 }-{ Q }_{ 1 } }{ { Q }_{ 3 }+{ Q }_{ 1 } } \)=\(\frac { 40-15 }{ 40+15 } =\frac { 25 }{ 55 } =0.455\)
Hence coefficient of QD = 0.455

Calculation of Harmonic Mean

\(HM=\frac { n }{ \sum { \left( \frac { f }{ X } \right) } } =\frac {120 }{ 5.975 } =20.08\)

In averaging ratios and percentages, geometric mean is more appropriate. Let us consider X represents Expenses at the end of the year.

GM = Anti \(\log { \left( \frac { \sum { logX } }{ N } \right) } \)
= Anti \(\log { \left( \frac { 6.4428 }{ 3 } \right) } \)
= Anti log(2.1476)
GM = 140.5
Average rate of increase in overhead expenses
140.5 – 100 = 40.5 %

P(A₁) = P(that the machine A₁ produces screws) = \(\frac{1000}{6000}=\frac{1}{6}\)
P(A₂) = P(that the machine A₂ produces screws) = \(\frac{2000}{6000}=\frac{1}{3}\)
P(A₃) = P(that the machine A₃ produces screws) = \(\frac{3000}{6000}=\frac{1}{2}\)
Let B be the event that the chosen screw is defective
\(\therefore\) P(B/A₁) = P(that defective screw from the machine A₁) = 0.01
P(B/A₂) = P(that defective screw from the machine A₂) = 0.015 and
P(B/A₃) = P(that defective screw from the machine A₃) = 0.02
We have to find P(A₁/B)
Hence by Baye’s theorem, we get
\(P(A_1/B)=\frac{P(A_1)P(B/A_1)}{P(A_1)P(B/A_1)+P(A_2)P(B/A_2)+P(A_3)P(B/A_3)}\)
\(=\frac{(\frac{1}{6})(0.01)}{(\frac{1}{6})(0.01)+(\frac{1}{3})(0.015)+(\frac{1}{2})(0.02)}\)
= \(\frac{0.01}{0.01+0.03+0.06}=\frac{0.01}{0.1}=\frac{1}{10}\)