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

Consider a 2,000 individuals getting injection of a given serum , n = 2000
Let X be the number of individuals suffering a bad reaction.
Let p be the probability that an individual suffers a bad reaction = 0.001
and q = 1– p = 1– 0.001 = 0.999
Since n is large and p is small, Binomial Distribtuion approximated to poisson distribution
So, λ = np = 2000 × 0.001 = 2
(i) Probability out of 2000, exactly 3 will suffer a bad reaction is
\(P(X=3)=\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } =\frac { { e }^{ -2 }{ 2 }^{ 3 } }{ 3! } =0.1804\)
(ii) Probability out of 2000, more than 2 individuals will suffer a bad reaction
= P(X > 2)
1-[P(X\(\le\)2)]
= 1 – [P(x = 0) + P(x = 1) + P(x = 2)]
\(=1-\left[ \frac { { e }^{ -2 }{ 2 }^{ 0 } }{ 0! } +\frac { { e }^{ -2 }{ 2 }^{ 1 } }{ 1! } +\frac { { e }^{ -2 }{ 2 }^{ 2 } }{ 2! } \right] \)
\(=1-{ e }^{ 2 }\left( \frac { { 2 }^{ 0 } }{ 0! } +\frac { { 2 }^{ 1 } }{ 1! } +\frac { { 2 }^{ 2 } }{ 2! } \right) \)
= 0.323

p = probability that a person will involve in an accident in a year
= 0.1/100 = 1/1000
given n = 10,000
so, \(\lambda\) = np = 10000\((\frac{1}{10000})\) = 10
Probability that not more than 5 will involve in such an accident in a year
P(X \(\le\) 5) = P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)+P(X = 5)
\(={ e }^{ -10 }[1+\frac { 10 }{ 1! } +\frac { { 10 }^{ 2 } }{ 2! } +\frac { { 10 }^{ 3 } }{ 3! } +\frac { { 10 }^{ 4 } }{ 4! } +\frac { { 10 }^{ 5 } }{ 5! } ]\)
= 0.06651

Probability of raining on a particular day is given by p = 9/30 = 3/10 and q = 1–p = 7/10.
The binomial distribution is P(X = x) = ⁿC_x p^x q^n–x
There are 7 days in a week
P(X = x) =\(\left( \begin{matrix} 7 \\ x \end{matrix} \right) { \left( \frac { 3 }{ 10 } \right) }^{ x }{ \left( \frac { 7 }{ 10 } \right) }^{ 7-x }\)
The probability of raining for atleast 2 days is given by
P(X\(\ge\)2) = 1-P(X<2)
= 1-[P(X = 0)+P(X = 1)]
Here, P(X = 0)\(\left( \begin{matrix} 7 \\ 0 \end{matrix} \right) { \left( \frac { 3 }{ 10 } \right) }^{ 0 }{ \left( \frac { 7 }{ 10 } \right) }^{ 7-0 }\)
= 0.0823
and \(P(X=1)=\left( \begin{matrix} 7 \\ 1 \end{matrix} \right) \left( \frac { 3 }{ 10 } \right) { \left( \frac { 7 }{ 10 } \right) }^{ 7-1 }\)
= 0.2471
Therefore the required probability = 1– [P(x = 0) +P(x = 1)]
= 1– [0.082+ 0.247]
= 0.6706