Let X be a random variable follows binomial distribution with p = q = 1/2
P (3 heads) = \(5{ C }_{ x }{ \left( \frac { 1 }{ 2 } \right) }^{ x }{ \left( \frac { 1 }{ 2 } \right) }^{ 5-x }\)
\(={ 5C }_{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 5-3 }\)
\(=5{ C }_{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }\)
\(=\frac { 5 }{ 16 } \)

We are given mean μ = 2.25 and standard deviation σ = 0.25.
Probability that weight of fish is less than 2 kg is P(X < 2.0)
When x = 20 \(Z=\frac { X-\mu }{ \sigma } =\frac { 2.0-2.25 }{ 0.25 } =P(Z<-1.0)=P(Z>1.0)\)
= 0.5 – 0.3413 = 0.1587
Therefore 15.87% of fishes weigh less than 2 kg.

Let the distribution of heights be normally distributed with mean mean 68.22 and standard deviation = 3.286
\(Z=\frac { X-\mu }{ \sigma } =\frac { X-69.25 }{ 3.286 } \)
When X = 74
\(Z=\frac { X-\mu }{ \sigma } =\frac { 74-69.25 }{ 3.286 } =1.4455\)
Now P(Z > 74) = P(Z > 1.44)
= 0.5 – 0.4251
= 0.0749
Expected number of children to be over 74 cm out of 1200 children
= 1200 × 0.0749 ≈ 90 children

Here p = q = 1/2
(a) probability of A beating B in exactly 3 games out of 4
\(\left( \begin{matrix} 4 \\ 3 \end{matrix} \right) { (\frac { 1 }{ 2 } ) }^{ 3 }\left( \frac { 1 }{ 2 } \right) ^{ 4-3 }\)
= 1/4 = 25%
(b) probability of A beating B in exactly 5 games out of 8
\(\left( \begin{matrix} 8 \\ 5 \end{matrix} \right) { (\frac { 1 }{ 2 } ) }^{ 5 }\left( \frac { 1 }{ 2 } \right) ^{ 8-5 }\)
\(=\frac { 7 }{ 32 } \) = 21.875%
Clearly, the first event is more probable.

The p.m.f. Binomial distribution is
p(x) = ⁿC_xp^xq^n-x
Given that E(x) = 2
For the Binomial distribution mean is given by np = 2 ... (1)
Given that var (x) = 4/3
For Binomial distribution variance is given by npq=4/3   ...(2)
\(\frac { (2) }{ (1) }⇒\frac {npq}{np} =\frac { \frac { 4 }{ 3 } }{ 2 } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
q = 2/3 and p = 1–2/3 = 1/3
Substitute is (1) we get
n = 6
Hence, P(X = 5) = 6C₅\({ \left( \frac { 1 }{ 3 } \right) }^{ 5 }{ \left( \frac { 2 }{ 3 } \right) }^{ 6-5 }\) = 0.0108

The parameters of Binomial distribution are n and p
For Binomial distribution Mean = np = 20
Standard deviation =\(\\ \\ \\ \sqrt { npq } =4\)   
∴ npq = 16
⇒ npq/np = 16/20 = 4/5
q = 4/5
⇒ p = 1–q = 1– (4/5) = 1/5
Since np = 20
n = \(\frac{20}{p}\)
n = 100

Let X denote the length of life of dry battery cells follows normal distribution with mean 12 and SD 3 hours

(i) more than 13 hours
P(X > 13)
When X = 13
\(Z=\frac { X-\mu }{ \sigma } =\frac { 13-12 }{ 3 } =0.333\)
P(X > 13) = P(Z > 0.333) = 0.5 – 0.1293 = 0.3707
The expected battery cells life to have more than 13 hours is 125 × 0.3707 = 46.34%

(ii) less than 5 hours
P(X < 5)
When X = 5
\(Z=\frac { X-\mu }{ \sigma } =\frac { 5-12 }{ 3 } =-2.333\)
P(X < 5) = P(Z < –2.333) = P(Z > 2.333)
= 0.5 – 0.4901 = 0.0099
The expected battery cells life to have more than 13 hours is 125 × 0.0099 = 1.23%

(iii) between 9 and 14 hours
When X = 9
\(Z=\frac { X-\mu }{ \sigma } =\frac { 9-12 }{ 3 } =-1\)
When X = 14
\(Z=\frac { X-\mu }{ \sigma } =\frac { 14-12 }{ 3 } =0.667\)
P(9 < X < 14) = P(–1 < Z < 0.667)
= P(0 < Z < 1) + P(0 < Z < 0.667)
= 0.3413 + 0.2486
= 0.5899
The expected battery cells life to have more than 13 hours is 125 x 0.5899 = 73.73%

Let X be the normal variate showing the score of the candidate with mean 45 and standard deviation 10.
(i) less than 35 marks
When X = 35
\(Z=\frac { X-\mu }{ \sigma } =\frac { 35-45 }{ 10 } =-1\)
P(X < 35) = P(Z < –1)
P(Z > 1) = 0.5 – P(0 < Z < 1)
= 0.5 – 0.3413
= 0.1587
Expected number of students scoring less than 35 marks are 0.1587 × 1300
= 206

(ii) more than 65 marks
When X = 65
\(Z=\frac { X-\mu }{ \sigma } =\frac { 65-45 }{ 10 } =2.0\)
P(X > 65) = P(Z > 2.0)
0.5 – P(0 < Z < 2.0)
0.5 – 0.4772
= 0.0228
Expected number of students scoring more than 65 marks are 0.0228 x 1300 = 30

Here mean μ= 30 and standard deviation σ = 5
(i) When X = 26 Z=(X−μ)/\(\sigma \) = (26 – 30)/5 = –0.8
And when X = 40 , z =\(\frac{40-30}{5}=2\)
Therefore,
P(26 < x < 40)) = P(–0.8 ≤ Z < 2)
= P(–0.8 ≤ Z ≤ 0) + p(0 ≤ Z ≤ 2)
= P(0 ≤ Z ≤ 0.8) + P(0 ≤ Z ≤ 2)
= 0.2881 + 0.4772 (By tables)
= 0.7653

(ii) The probability that X≥45
When X = 45
\(z=\frac { X-\mu }{ \sigma } =\frac { 45-30 }{ 5 } =3\)
P( X ≥ 45) = P(Z ≥ 3)
= 0.5 – 0.49865
= 0.00135

For Binomial Distribution the mean is np and varaiance is npq
Given values are np + npq = 24 np(1 + q) = 24 – (1)
Other term np × npq = 128  n²p²q = 128-(2)
From (1) we get np = 24/(1+q) which implies n²p² = (24/(1+q))²
Substitute this value in equation (2) we get
\({ \left( \frac { 24 }{ 1+q } \right) }^{ 2 }q=128\)   which implies 9q = 2(1+2q+q²)
(2q – 1)(q – 2) = 0
Where q =\(\frac{1}{2}\) and p =\(\frac{1}{2}\)
Substitute in (1) we get n = 32
Hence the binomial distribution \({ 32C }_{ x }{ \left( \frac { 1 }{ 2 } \right) }^{ x }{ \left( \frac { 1 }{ 2 } \right) }^{ 32-x }\)