Let X denote the height of the student
Given P (X < 40) = 10% = \(\frac { 10 }{ 100 } \) =0.1
P(X> 90) = 10% = \(\frac { 10 }{ 100 } \) =0.1

∴ P(40 < X < 90) = P(-∞ < X < ∞) - [P(X < 40) + P(X < 90)]
= 1 - (0.1 + 0.1)
= 1 - 0.2 = 0.8
∴ out of 800 students, number of students scored between 40 and 90 = 800 x 0.8
= 640 students.

Let X denote the height of the student
Given μ = 64.5 inches and σ = 3.3 inches
Given that P(-∞ < Z < C) = 0.99
⇒ P( -∞ < Z < 0) + P (0 < Z < C) = 0.99
⇒ 0.5 + P (0 < Z < C) = 0.99
⇒ P(0 < Z < C) = 0.99 - 0.5 = 0.49...(1)
From the standard normal distribution table
P(0 < Z < 2.33) = 0.49...(2)
From (1) & (2), C = 2.33
we know that Z = \(\frac { X-\mu }{ \sigma } \)
⇒ 2.33 = \(\frac { X-64.5 }{ 3.3 } \)
⇒ X = (2.33) (3.3) + 64.5
⇒ X = 72.19 inches
Hence, the height below which 99% of the student lie is 72.19 inches.

Let X denotes the weight of the male students
Given μ = 151, σ = 15 and N = 500
(i) between 120 and 155 pounds
P(120 < X < 155)
When X = 120, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 120-151 }{ 15 } \)
= \(\frac { -31 }{ 15 } \) = -2.067
When x = 155, Z = \(\frac { 155-151 }{ 15 } \)
= \(\frac { 4 }{ 15 } \) = 0.2667

∴ P(120< X < 155) = P(-2.067 < X < 0.2667)
= P (-2.067 < Z < 0) + P (0 < Z < 0.2667)
= P (0 < Z < 2.067) + P (0 < Z < 0.2667)
(By symmetry)
= 0.4803 + 0.1026 = 0.5829
Probability for a student weigh between 120 and 155 pounds is 0.5829
∴ Out of 500 students, number of students weighing between 120 and 155 pounds
= 500 \(\times\) 0.5829 = 291 students
(ii) more than 185 pounds
P(X > 185)
When X = 185, Z = \(\frac { 185-151 }{ 15 } \)
=\(\frac { 34 }{ 15 } \) = 2.2667

∴ P(X >185) = P (Z > 2.2667)
= P (2.2667 < Z < ∞)
= P(0 < Z < ∞) - (0 < Z < 2.2667)
= 0.5 - 0.4881
= 0.0119
i.e Probability for a student weighing above 185 is 0.0119
∴ out of 500 male students, number of students weighing more than 185 pounds.
= 500 \(\times\) 0.0119 = 6 students.

Given n = 10, p = \(\frac { 20 }{ 100 } =\frac { 1 }{ 5 } \)
∴ q = 1-p = \(1-\frac { 1 }{ 5 } =\frac { 4 }{ 5 } \)
Let
X denote the number of defective bolts chosen
∴ X = 2
(i) Using binomial distribution
P(X = 2) = \(10{ C }_{ 2 }\left( \frac { 1 }{ 5 } \right) ^{ 2 }\left( \frac { 4 }{ 5 } \right) ^{ 8 }\)
= \(\frac { 10\times 9 }{ 2\times 1 } \left( \frac { { 4 }^{ 8 } }{ { 5 }^{ 10 } } \right) =45\left( \frac { 4^{ 8 } }{ { 5 }^{ 10 } } \right) \)
(ii) Using Poisson distribution
λ = np = 10 \(\times\) \(\frac { 1 }{ 5 } \) = 2
P(X = x) \(\times\)\(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
x = 0,1,2,......n
∴ P(X = 2) = \(\frac { e^{ -2 }(2^{ 2 }) }{ 2 } =e^{ -2 }\left( \frac { 4 }{ 2 } \right) \)
= 2e^-2
= 2(0.1353) = 0.2706
∴ P(X = 2) = 0.2706

Given n = 4
p = probability of getting a head when a coin is tossed
= \(\frac { 1 }{ 2 } \)
∴ q=1-p =\(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
X is a ran dom verticical having
p(X = x) = nC_x p^xq^n-x, x = 0,1,2,3,4
a) atleast 2 heads
P(X ≥ 2) = 1-P(X<2)
= 1- [P(X = 0) + P(X = 1)]
=1-\(\left[ 4C_{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 4 }+4C_{ 1 }\left( \frac { 1 }{ 2 } \right) ^{ 1 }\left( \frac { 1 }{ 2 } \right) ^{ 3 } \right] \)
=1-\(\left[ 1\left( \frac { 1 }{ 16 } \right) +4\left( \frac { 1 }{ 8 } \right) \left( \frac { 1 }{ 2 } \right) \right] \)
=1-\(1-\left( \frac { 4 }{ 16 } +\frac { 1 }{ 16 } \right) =1-\frac { 5 }{ 16 } =\frac { 11 }{ 16 } \)
P(X ≥ 2) = \(\frac { 11 }{ 16 } \)
b) atmost 2 heads
P(X≤2) = 1-P(X>2)
= 1 - [P(X = 3) + P (X = 4)]
= 1-\(\left[ { 4C }_{ 3 }\left( \frac { 1 }{ 2 } \right) ^{ 3 }\left( \frac { 1 }{ 2 } \right) ^{ 1 }+{ 4C }_{ 4 }\left( \frac { 1 }{ 2 } \right) ^{ 4 }\left( \frac { 1 }{ 2 } \right) ^{ 0 } \right] \)
= 1-\(\left[ 4\left( \frac { 1 }{ 8 } \right) \left( \frac { 1 }{ 2 } \right) +1\left( \frac { 1 }{ 16 } \right) \right] \)
= \(1-\left( \frac { 4 }{ 16 } +\frac { 1 }{ 16 } \right) =1-\frac { 5 }{ 16 } =\frac { 11 }{ 16 } \)
P(X≤2) =\(\frac { 11 }{ 16 } \).