\( \mathrm{P}(\mathrm{X} =\mathrm{x})=n \mathrm{C}_{x} p^{x} q^{n-x} \)
\(\mathrm{P}(\mathrm{X} =3)=2 \mathrm{P}(\mathrm{x}=2) \)
\(5 \mathrm{C}_{3} \mathrm{p}^{3} \mathrm{q}^{2} =2\left(5 \mathrm{C}_{2} \mathrm{p}^{2} \mathrm{q}^{3}\right) \)
\(\mathrm{p} =2 \mathrm{q} \)
\(\mathrm{p} =2(1-\mathrm{p}) \Rightarrow 3 \mathrm{p}=2 \Rightarrow \mathrm{p}=\frac{2}{3}\)

Let λ be the parameter
Given 3. P(X = 2) = P(X = 4)
⇒ 3. \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =\frac { e^{ -\lambda }.{ \lambda }^{ 4 } }{ 4! } \)
\(\frac { 3{ \lambda }^{ 2 } }{ 2 } =\frac { { \lambda }^{ 4 } }{ 4\times 3\times 2 } \)
⇒ 36λ² = λ⁴
λ⁴-36λ² = 0
⇒ λ²(λ²-36) = 0
⇒ λ² = 0 or λ² = 36
⇒ λ = 6 since λ > 0

The standard deviation of the sampling distribution of a statistic is known as its Standard Error.

Givens \(\sigma^2\) = 4 which implies \(\sigma\) = 2, n = 1 hour = 60 min, \(\bar { X } \) = 20/min
Standard Error \(=\frac { \sigma }{ \sqrt { n } } =\frac { 2 }{ \sqrt { 60 } } =0.2582\)

Given mean of a binomial distribution is 5
np = 5 ....(1)
Also, standard deviation is 2 ⇒ Variance = 2² = 4
∴ npq = 4...(2)
(2) \(\div \) (1) given, \(\frac { npq }{ np } \)=\(\frac { 4 }{ 5 } \)
⇒ q = \(\frac { 4 }{ 5 } \)
p = 1-q = \(1-\frac { 4 }{ 5 } =\frac { 1 }{ 5 } \)
Substituting p = \(\frac { 1 }{ 5 } \) in (1) we get
n\(\left( \frac { 1 }{ 5 } \right) \) = 5 ⇒ n = 25
∴ The binomial distribution is ⁿC_x p^xq^n-x, x = 0,1,2,....n
⇒ ²⁵C_x \(\left( \frac { 1 }{ 5 } \right) ^{ x }\left( \frac { 4 }{ 5 } \right) ^{ 25-x }\), x = 0,1,2,....25.

The average number of typographical errors per page in the book is given by \(\lambda\) = (390/520) = 0.75.
Hence using Poisson probability law, the probability of x errors per page is given by
\(P(X=x)=\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } ={ e }^{ -0.75 }=\frac{(0.75)^x}{x!}\) x = 0,1,2,3……
The required probability that a random sample of 5 pages will contain no error is given by :
[P(X = 0)]⁵ = (e-0.75)⁵ = e^-3.75

Let X be a random variable follows binomial distribution with probability value p = 1/2 and q = 1/2
probability that exactcy 2 heads occur are as follows
P(X = 2) = \(\left( \begin{matrix} 6 \\ x \end{matrix} \right) { p }^{ x }{ q }^{ n-x }\)
\(=\left( \begin{matrix} 6 \\ 2 \end{matrix} \right) { \left( \frac { 1 }{ 2 } \right) }^{ 2 }{ { \left( \frac { 1 }{ 2 } \right) } }^{ 6-2 }\)
\(=\frac { 15 }{ 64 } \)

Null hypothesis :
"Null hypothesis is the hypothesis which is tested for possible rejection under the assumption that it is true", and it is denoted by H_o.
For example : If we want to find the population mean has a specified value \(\mu _o\) then the null hypothesis H_o is set as follows \(H_{0}: \mu=\mu_{0}\)
Critical region or Rejection region :
A region corresponding to a test statistic in the sample space which tends to rejection of H_o is called critical region or region of rejection.

Let x be the random variable denoting the number of heads
\( \mathrm{p} =\frac{1}{2}, \mathrm{q}=\frac{1}{2}, \mathrm{n}=6 \)
\(\mathrm{P}(\mathrm{X} =\mathrm{x})=n C_{x} p^{x} q^{x-x} \)
\(\mathrm{P}(\mathrm{X} \geq 4) =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6) \)
\( =6 C_{4}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{2}+6 C_{5}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{1}+6 C_{6}\left(\frac{1}{2}\right)^{6} \)
\( =\frac{1}{2^{6}}\left[6 C_{2}+6 C_{1}+6 C_{0}\right]=\frac{11}{32}=0.34375\)

Sample size n = 50
Sample S.D s = 6.3
Population S.D \(\sigma\) = 6
The standard error for sample S.D is given by 
\(S.E=\sqrt { \frac { { \sigma }^{ 2 } }{ 2n } } =\frac { 6 }{ \sqrt { 2(50) } } =\frac { 6 }{ \sqrt { 100 } } =0.6\)
Thus standard error for sample S.D = 0.6.

If the occurrence of 3 or 4 on the die is called a success, then
Sample size = 9000; Number of Success = 3240
Sample proportion =\(P=\frac { 3240 }{ 9000 } =0.36\)
Population proportion (P) = Prob(getting 3 or 4 when a die is thrown)
\(=\frac { 1 }{ 6 } +\frac { 1 }{ 6 } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } =0.3333\)
Thus P = 0.3333 and Q = 1–P = 1– 0.3333 = 0.6667
The S.E for sample proportion is given by
\(S.E=\sqrt { \frac { PQ }{ N } } =\sqrt { \frac { (0.3333)(0.6667) }{ 9000 } } =0.00496\)
Hence the standard error for sample proportion is S.E = 0.00496.

(i) Binomial distribution is symmetrical if p = q = 0.5. It is skew symmetric if p≠q. It is positively skewed if p < 0.5 and it is negatively skewed if p > 0.5.
(ii) For binomial distribution, variance is less than mean.
Variance = npq = (np)q < np < mean.

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

Probability of getting a degree p = 0.4
∴ q = 1– p
= 1 - 0.4
= 0.6
(i) P (one will be a graduate) = P(X = 1) = 5C₁ (0.4)(0.6)⁴
= 0.2592
(ii) P ( atleast one will be a graduate) = 1–P (none will be a graduate)
= 1-5C₀(P⁰)(Q)5-0
= 1-5C₀(0.4)⁰(0.6)5
= 1-0.0777
= 0.9222

\( \mathrm{Z}_{\mathrm{i}} =\frac{x_{2}-\mu}{\sigma} \)
\(0.84 =\frac{x_{2}-40}{25} \)
\(x_{2} =61 \)
\(\mathrm{Z}_{2} =\frac{x_{1}-\mu}{\sigma} \)
\(-0.84 =\frac{x_{i}-40}{25} \)
\(x_{2} =19 \)
60% of students score between 19 and 61.

Given Sample size n = 400
Sample mean \(\\ \bar { X } =67.47\)inches
Population mean μ = 67.39 & σ = 1.30 inches)
Null Hypotheses H_o:
μ = 67.39 inches (i.e., the sample has been drawn from the population with μ = 67.39 & σ = 1.30 inches)
Alternative Hypotheses H₁:
μ ≠ 67.39 inches (two tail test)
(i.e., the sample has not been drawn from the population with μ = 67.39 & σ = 1.30 inches) The level of significance a = 5% = 0.05
Applying the test statistic,
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 67.47-67.39 }{ \frac { 1.30 }{ \sqrt { 400 } } } =\frac { 0.08 }{ 0.065 } =1.2308\)
\(\therefore |Z|=1.2308\)
The significant value \({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Here \(Z={ Z }_{ \frac { \alpha }{ 2 } }i.e.,1.2308<1.96\)
Inference: Since \({ Z }_{ \frac { \alpha }{ 2 } }\)at 5% level of significance the null hypothesis H_o is accepted.
Hence, we conclude that the sample has been drawn from the population with mean height 67.39 inches and standard deviation 1.30 inches.

Sample size n = 400
Probability of vegetarian = \(\frac{230}{400}=0.575\)
∴ P = 0.575
Probability of non-vegetarian q = 1 - P = 1 - 0.575
⇒ q = 0.425
Standard error for sample proportion =\(\\ \\ \sqrt { \frac { pq }{ n } } \)
\(\sqrt { \frac { (0.575)(0.425) }{ 400 } } \)
\(=\sqrt { 0.000610 } \)
= 0.0246
S.E = 0.025(app)

Given, sample size = 100, \(\bar x\) = 7.4, since σ is unknown but s = 1.2 is known.
In this problem, we consider \(\breve { \sigma } =s\quad { Z }_{ \frac { \alpha }{ 2 } }=1.96\)
\(S.E=\frac { \breve { \sigma } }{ \sqrt { n } } =\frac { s }{ \sqrt { n } } =\frac { 1.2 }{ \sqrt { 100 } } =0.12\)
Hence 95% confidence limits for the population mean are
\(\bar { x } -{ Z }_{ \frac { \alpha }{ 2 } }\frac { \sigma }{ \sqrt { n } } <\mu <\bar { x } +{ Z }_{ \frac { \alpha }{ 2 } }\frac { \sigma }{ \sqrt { n } } \)
\(7.4-(1.96\times 0.12)\le \mu \le 7.4+(1.96\times 0.12)\)
\(7.4-0.2352\le \mu \le 7.4+0.2352\)
\(7.165\le \mu \le 7.635\)
This implies that the probability that the true value of the population mean breaking strength of the cotton threads will fall in this interval (7.165,7.635) at 95%.

Given μ = 68.0, σ = 3.0
(a) P(X>72)
When X = 72, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 72-68 }{ 3 } =\frac { 4 }{ 3 } \) = 1.33
∴ P(X>72) = P(Z>1.33)
= P(0 = 0.5-0.4082 = 0.0918

∴ Prob of 1 student having height more than 72 inches is 0.0918.
Out of 500 students, number of students having height more than 72 inches is 0.0918 x 500 = 45.9
= 0.46 (approximately).
(b) P(X ≤ 64)
When X = 64, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 64-68 }{ 3 } =\frac { -4 }{ 3 } \) = -1.33
∴ P(X≤64) = P(Z≤-1.33)
= P(-∞ = 0.5-P(0 = 0.5-0.4082
= 0.0918

∴ Probability of 1 student having height less than 64 inches is 0.0918
∴ Out of 500 students, number of students having height less than 64 inches is = 0.0918 x 500 = 45.9 = 46 (approximately).
(c) P(65 When X = 65, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 65-68 }{ 3 } =\frac { 3 }{ 3 } \) = 1
When X = 71, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 71-68 }{ 3 } =\frac { 3 }{ 3 } \) = 1
∴ P(65 = P(-1 = P(0 [By symmetry]
= 2. P(0 = 2(0.3413) = 0.6826)

∴ Probability of 1 student having height between 65 and 71 inches is 0.6826
∴ Out of 500 students, number of students having height between 65 and 71 inches is 0.6826 x 500 = 341.37
= 342 (approximately)

Let the probability for boy or girl is p =\(\frac { 1 }{ 2 } \) and q = \(\frac { 1 }{ 2 } \), n = 4
(i) P (atleast one boy) = P(X≥1)
= 1-P(X<1)
= 1-[P(X = 0)]
= 1-\(\left[ { { 4 }_{ C } }_{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 4 } \right] \)  p(x) = ⁿC_x p^xq^n-x, n = 4, x = 0
= \(\left( \frac { 1 }{ 2 } \right) ^{ 4 }=1-\frac { 1 }{ 16 } \) = 1-.0625
= 0.9375
∴ Probability of one family having atleast one boy = 0.9375.
Hence, probability of 750 families having atleast one boy
= 0.9375 \(\times\) 750 = 703
(ii) P (atmost 2 girls) = P(X ≤ 2)
= P(X = 0) + P(X = 1) + P(X = 2)
= ⁴C₀ \(\left( \frac { 1 }{ 2 } \right) ^{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 4 }\) + ⁴C₁ \(\left( \frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 2 } \right) ^{ 3 }\) + ⁴C₂ \(\left( \frac { 1 }{ 2 } \right) ^{ 2 }\left( \frac { 1 }{ 2 } \right) ^{ 2 }\)
= \(\left( \frac { 1 }{ 2 } \right) ^{ 4 }\) [⁴C₀ + ⁴C₁ + ⁴C₁]
= \(\frac { 1 }{ 16 } \left[ 1+4+\frac { 4\times 3 }{ 2\times 1 } \right] \) = 0.0625 (5+6)
= 0.0625 \(\times\) 11 = 0.6815
∴ Probability of one family having atmost 2 girls = 0.6875.
Hence, probability of 750 families having atmost 2 girls
= 0.6875 \(\times\) 750 = 515.6 = 516.
(iii) P(Chiidren of both sexes) = P (atleast one girl and atleast one boy)
= 1[P(X = 0) + P(X = 4)]
= 1-⁴C₀ \(\left( \frac { 1 }{ 2 } \right) ^{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 4 }-{ C }_{ 4 }\left( -\frac { 1 }{ 2 } \right) ^{ 4 }\left( \frac { 1 }{ 2 } \right) ^{ 0 }\)
= \(1-\frac { 1 }{ 16 } -\frac { 1 }{ 16 } =1-\frac { 2 }{ 16 } =1-\frac { 1 }{ 8 } =\frac { 7 }{ 8 } \)
∴ Probability of 1 family of having children of both sexes = 0.875.
∴ Probability of 750 families having children of both sexes = 0.875 \(\times\) 75 = 656.

(i) P (Exactly one defective bolt)
= P(X = 1)
= ⁴C₁ (0.18)¹ (0.82)³     [p(x) = ⁿC_x p^xq^n-x, n = 4, x = 1]
= 4(0.18) (0.82)³
= (0.72) (0.5513)
P(X = 1) = 0.3969
(ii) P (none will be defective)
= P(X = 0)
=⁴C₀ (0.18)⁰ (0.82)⁴     [p(x) = ⁿC_x p^xq^n-x, n = 4, x = 0]
= (0.82)⁴
[⁴C₀ = 1 and (0.18)⁰ = 1]
P(X = 0) = 0.4521.
(iii) P (atmost 2 will be defective)
= P(X≤2)
= P(X = 0) + P(X = 1) + P(X = 2)
= 0.4521 + 0.3969 + ⁴C₂ (0.18)² (0.82)²
= 0.849 + 6(0.18)² (0.82)²
= 0.849 + 0.1307 = 0.9797.