Given μ = 30, σ = 4
(a) P(X < 40)
When X = 40, Z = \(\frac { X-\mu }{ \sigma } \)
=\(\frac { 40-30 }{ 4 } =\frac { 10 }{ 4 } \) = 2.5
∴ P(X<40) = P(Z<2.5)

= P(-∞) = 0.5+.4938
P(X<40) = 0.9938
(b) P(X>21)
When X = 21, Z = \(\frac { 21-30 }{ 4 } \quad \)
= \(\frac { -9 }{ 4 } \) = -2.25
∴ P(X>21) = P(Z>-2.25)

= 0.4878 + 0.5
P(X > 21) = 0.9878
(c) P(30 When X = 30, Z₁ =\(\frac { 30-30 }{ 4 } \) = 0
When X = 35, Z₂ = \(\frac { 35-30 }{ 4 } \) = 1.25
P(30 = 0.3944

(a) P(X<40,000)
When X = 40,000.
Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 40,000-50,000 }{ 20,000 } \) = -0.5
∴ P(X<40,000) = P(Z<-0.5)
= P(-∞ =P(-∞ < Z <0) - P(-0.5 = 0.5-P(0 [By symmetry]
= 0.5-0.1915 = 0.3085
∴ 30.85 % pf people earn less than $40,000

(b) P(45,000 < X < 65,000)
When X = 45,000
Z₁ = \(\frac { 45,000-50,000 }{ 20,000 } \) = -0.25
When X=65,000
Z=\(\frac { 65,000-50,000 }{ 20,000 } \)=0.75
∴ P(45000 < X < 65,000) = P(-0.25 < Z < 0.75)

= P (-0.25 < Z < 0) + P(0 < Z < 0.75)
= P(0 < Z < 0.25) + P (0 < Z < 0.75)
= 0.0987 + 0.2734 = 0.3721
Hence, 37.21% of people earn between Dallor 45,000 and Dallor 65,000.
(c) P(X>7000)
When X = 7000
Z = \(\frac { 75000-50,000 }{ 20,000 } \) = 1.25
∴ P(X > 7000) = P(Z > 1.25)
= 0.5 - 0.3944
= 0.1056

 Hence, 10.56% of people earn more than Dallor 75000.

Given μ = 20, σ = 2
a) P(X<19.5)
When X = 19.5, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 19.5-20 }{ 20 } \) = -0.25
∴ P(Z<-0.25) = P(-∞ =P(-∞

= 0.5-P(0 = 0.5-0.0987
P(X<19.5) = 0.4013
b) P(between 20 and 22 hours)
= P(20 When X = 20, Z₁ = \(\frac { 20-20 }{ 2 } \) = 0
When X = 22, Z₂ = \(\frac { 22-20 }{ 2 } \) = 1

∴ P (20 < X < 22) - P(0 < Z < 1)
∴ P (20 < X < 22) = 0.3413

Given λ average = 300 per hour
= \(\frac { 300 }{ 60 } \) per minutes
∴ λ = 5
∴ X follows a poisson distribution
(i) P (none passes in a minute) = P (X = 0)
= \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } \)
[∵ P(x, λ) = \(\frac { { e }^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)]
P(X=0) = \(\frac { { e }^{ -5 }.{ (-5) }^{ 0 } }{ 0! } \)
= e^-5 = 0.0067379
(ii) P(X = 2) = \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =\frac { { e }^{ -5 }.(5)^{ 2 } }{ 2 } \)
= \(\frac { (0.0067379)(25) }{ 2 } \)
= 0.08422
∴ Probability of expected number passing in 2 minutes = .08422

Let p be the probability getting his piston rejected
Given p = 12% = \(\frac { 12 }{ 100 } \) = 0.12
∴ q = 1-p = 1-0.12 = 0.88
n = 10
(a) P (not more than 2 rejects)
= P(X≤2) = P(X = 0) + P(X = 1) + P(X = 2)
= 10C₀(0.12)⁰ (0.88)¹⁰ + 10C₁ (0.12)¹ (0.88)⁹ + 10C₂ (0.12)² (0.88)⁸
[∵ P(x) = nC_x p^xq^n-x]

= (0.88)⁸ [(0.88)² + 10(0.12) (0.88) + 45 (0.12)²]
= (0.88)⁸ [0.7744 + 1.056 + 0.648]
= (0.3596) (2.4784) = 0.8913
∴ Probability of not more than 2 rejects = 0.8913
b) P(at least 2 rejects)
= P(X ≥ 2) = 1 - P (X < 2)
= 1 - [P(X = 0) + P (X = 1)]
= 1-[10C₀(0.12)⁰ (0.88)¹⁰ + 10C₁ (0.12)¹ (0.88)⁹ ]
= 1 - [(0.88)¹⁰ + 10 (0.12) (0.88)⁹]
= 1 - (0.88)⁹ [0.88 + 1.2] = 1 - (0.31647) (2.08)
= 1 - 0.6583 = 0.34173
P (atleast 2 rejects) = 0.34173

Given μ = 400, σ = 100
(i) Company pays a penalty of atleast Rs. 2,00,000
Penalty per day = Rs.10,000
Number of days = \(\frac { 2,00,000 }{ 10,000 } \) = 20
Hence, the company has taken excess of 20 days
∴ P(atleast 470 days)     [∵ 450 + 20 = 470]
= P(X ≥ 470)
When X = 470, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 470-400 }{ 100 } =\frac { 70 }{ 100 } \) = 0.7
∴ P(X ≥ 470) = P(Z ≥ 0.7)
= P(0.7 = 0.5-0.2580
P(X ≥ 470) = 0.2420

(ii) P (atmost 500 days) = P(X≤500)
When X = 500, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 500-400 }{ 100 } =\frac { 100 }{ 100 } \)=1
∴ P(X ≤ 500) = P(Z≤1)
= P(-∞ = 0.5+0.3413
= 0.8413
P(X ≤ 500) = 0.8413

Given μ = 16.28 seconds and σ = 0.12 second
P(X<16.35)
When X = 16.35, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 16.35-16.28 }{ 0.12 } \)
= \(\frac { 0.007 }{ 0.12 } \) = 0.583
∴ P(X < 16.35) = (Z < 0.583)
= P(-∞  = 0.5 + 0.2190 = 0.7190

∴ Probability that it will take less than 16.35 sec to develop prints = 0.719.

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)

Given μ = 12 and σ = 4
(1) P( ≤ 20)
When X = 20, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 20-12 }{ 4 } =\frac { 8 }{ 4 } \) = 2
∴ P(X≤20) = P(Z≤2)
= P(-∞ = 0.5+0.4772
P(X≤20) = 0.9772

(ii) P(0≤X≤12)
When X=0, Z=\(\frac { X-\mu }{ \sigma } \)
\(\frac { 0-12 }{ 4 } =\frac { -12 }{ 4 } \)=-3
When X=12, Z=\(\frac { X-\mu }{ \sigma } \)
=\(\frac { 12-12 }{ 4 } =\frac { 0 }{ 4 } \)=0
∴ P(0≤X≤12) = P(-3≤Z≤0)
= P(0≤Z≤3) (By symmetry)
P(0≤X≤12) = 0.4987.

Plot the variable X = 50 on the left side and
X = 86 on the right side of the curve.
Given P( -∞< Z₁ < -Z₁) = 0.3
⇒ P(-Z₁ < Z < Z₁) = 0.2       [∵ 0.5 - 0.3 = 0.2]
⇒ P(0 < Z < Z₁) = 0.2 [By symmetry]
⇒ Z₁ = -0.52 [From the normal distribution table 'and it lies on the negative side]
⇒ -0.52 = \(\frac { X-\mu }{ \sigma } \) ⇒ -0.52 σ = 50-μ
⇒ 50-μ = -0.52 σ ....(1)
Also given P(Z₂ ∴ P(Z₂2) = 0.4    [∵ 0.5-0.1 = 0.4]
⇒ Z₂ = 1.28 (from the table)
⇒ 1.28 =\(\frac { 86-\mu }{ \sigma } \)
⇒ 86-μ = 1.28 σ ...(2)

Substituting σ = 20 in (2) we get,
86-μ = (1.28)(20)
86-μ = 25.6
μ = 86-25.6
μ = 60.4

Hence, the mean is 60.4 and standard deviation is 20.

Given mean μ = 2040 and σ = 60.
Let X denote the number of bulbs likely to burn.
(i) P(X > 2150)
When X = 2150, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 1950-2040 }{ 50 } \) = \(\frac { 110 }{ 60 } \) = 1.833
∴ P(X>2150) = P(Z>1.833)
= P(0 = 0.5-0.4664 [from std normal table]
= 0.0336.

∴ Probability of 1 bulb likely to burn for more than 2150 hours is 0.0336.
∴ Out of 2000 bulbs, number of bulbs likely to burn for more than 2150 hours is 0.0336 x 2000.
= 67.2
(ii) P (less than 1950 hours) = P( X < 1950)
when X = 1950, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 1950-2040 }{ 50 } \) = -1.5
∴ P(X<1950) = P(Z<-1.5)
= P(-∞ = 0.5-P(0 = 0.5-0.4332 = 0.0668

∴ Probability of 1bulb likely to burn forless than 1950 hours is 0.0668.
∴ Out of 2000 bulbs, number of bulbs likely to burn for less than 1950 hours is 0.0668 x 2000.
= 133.6
(iii) P (more than 1920 hours but less than 2100 hours)
= P(1920 < X < 2100)
When X = 1920,
Z = \(\frac { 1920-2040 }{ 60 } \) = -2
When X = 2100
Z = \(\frac { 2100-2040 }{ 60 } \) = 1
∴ P(1920 = P(-2 = P(0 (By symmetry)
= 0.4772 + 0.3413
= 0.8185

∴ Probability of 1 bulb likely to burn between 1920 and 2100 hours is 0.8185.
∴ Out of 2000 bulbs, number of bulbs likely to burn more than 1920 hours and less than 2100 hours is 0.8185 x 2000.
= 1637

(i) The curve is bell shaped and symmetrical about the line x = μ.
(ii) Mean, median and mode of the distribution coincide.
(lii) X-axis is an asymptote to the curve.
(iv) The points of inflexion of the curve are x = μ ± σ.
(v) The curve is unimodal.

Given average number of customers, appear in a counter
λ = 2
∴ X follows a poisson distribution with
p(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
(i) P (no customer appears)
= P(X = 0) = \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } \) = e^-λ = e^-2 = 0.1353   [∵ e^-2 = 0.1353]
(ii) P(3 or more customer appear)
= P(X≥3)
= 1-P(X<3)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= 1-\(\left[ \frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } \right] \)
= 1-e^-λ (1+λ+\(\frac { \lambda ^{ 2 } }{ 2 } \))
= 1-e^-2 (1+2+\(\frac { 4 }{ 2 } \))
= 1-e^-2 (5) = 1 - 0.1353(5)
= 1 - 0.6765 = 0.3235
Hence, probability of three or more customers appear in a counter of a certain bank is 0.3235.

Given mean = 4
∴ λ = 4
X follows Poisson distribution with p(x, λ)
= \(\frac { { e }^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
P (no accident per day) = P(X = 0)
(i) P(X = 0) = \(\frac { { e }^{ -4 }.(4)^{ 0 } }{ 0! } \)
= e^-4 = 0.0183   [∵ e^-4 = 0.0183]
∴ Probability of no accident in 1 day = 0.0183
∴ Probability of no accidents in 100 days
= 0.0183 \(\times\) 100 = 1.83 days
= 2 days (approximately)
(ii) P (atleast 2 accidents per day) = P(X≥2)
P(X≥2) = 1-P(X<2)
=1 - [P(X = 0) + P(X = 1)]
=1-\(\left[ \frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } \right] \)
= 1-e^-λ [1+λ]
= 1-e^-4 (1+4) =1-e^-4 (5)
= 1 - 0.0183(5) = 1 - 0.0915
= 0.0985.
∴ Probability of atleast 2 accidents in 1 day
= 0.9085
Hence, probability of atleast 2 accidents in 100 days
= 0.9085 x 100
= 90.85 days
= 91 days (approximately)
(iii) P (atmost 3 accidents per day)
= P(X≤3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= \(\left[ \frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 3 } }{ 3! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 4 } }{ 4! } \right] \)
= e^-λ \(\left( 1+\lambda +\frac { { \lambda }^{ 2 } }{ 2 } +\frac { { \lambda }^{ 3 } }{ 6 } \right) \)
= e^-4 (1+4+\(\frac { 16 }{ 2 } +\frac { 64 }{ 6 } \))
= e^-4 (1+4+8+\(\frac { 32 }{ 3 } \))
= 0.0183 (23.666) = 0.4331
∴ Probability of atmost 3 accidents in 1 day = 0.4331
Hence, probability of atmost 3 accidents in 100 days = 43.31
= 43 days (approximately).

Given average number of phone calls between 10 am and 2.30,pm per minute is 2.5.
∴ λ = 2.5
Hence X follows a Poisson distribution with
p(x, λ) = \(\frac { { e }^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
(i) P (no phone at all) = P(X = 0)
P(X = 0) = \(\frac { { e }^{ -2.5 }.(2.5)^{ 0 } }{ 0! } \) = e^-2.5
= 0.08208   [∵ e^-2.5 = 0.08208]
(ii) P (exactly 3 calls)
P(X = 3) = \(\frac { { e }^{ -2.5 }.(2.5)^{ 3 } }{ 3! } \)  [∵ λ = 2.5 and x=3]
= \(\frac { (0.08208)(2.5)^{ 3 } }{ 3\times 2 } \) = 0.21375
∴ P(X = 3) = 0.2138
(iii) P (atleast 5 calls) = P(X ≥ 5)
P(X≥5) = 1-P(X<5)
=1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)]
= 1-\(\left[ \frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 3 } }{ 3! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 4 } }{ 4! } \right] \)
= 1-e^-λ \(\left( 1+\lambda +\frac { { \lambda }^{ 2 } }{ 2 } +\frac { { \lambda }^{ 3 } }{ 6 } +\frac { { \lambda }^{ 4 } }{ 24 } \right) \)
= 1-e^-2.5 (1+2.5+\(\frac { ({ 2.5) }^{ 2 } }{ 2 } +\frac { (2.5)^{ 3 } }{ 6 } +\frac { (2.5)^{ 4 } }{ 24 } \))
= 1-e^-2.5 (1 + 2.5 + 3.125 + 2.604) + 1.6276)
= 1-e^-2.5 (10.8566)
= 1 - 0.08208 (10.8566) = 1 - 0.89111
P(X ≥ 5) = 0.1089.

Given mean = λ = 1.5
X follows poisson distribution with
p(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
(i) ∴ P(neither car is used)
= P(X = 0) = \(\frac { e^{ -1.5 }.(1.5)^{ 0 } }{ 0! } \) =e^-1.5
= 0.2231    [∵ e^-1.5 =0.2231]
(ii) P (Some demand is refused)
The demand may be either 0 car, 1 car or 2 cars
∴ P (Some demand is refused) = 1 - P(X ≤ 2)
= 1 - [P(X = 0) + P (X = 1) + P(X = 2)]
= 1-\(\left[ \frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { e^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } +\frac { e^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } \right] \)
= 1-e^-λ (1+λ+\(\frac { { \lambda }^{ 2 } }{ 2 } \))
= 1-e^-1.5 (1+1.5+\(\frac { (1.5)^{ 2 } }{ 2 } \))
= 1 - 0.2231 (3.625) = 1 - 0.8087 = 0.1912
∴ Probability of some demand is refused = 0.1912.

Mean E(X) =\(\overset { \infty }{ \underset { x=0 }{ \Sigma } } x.p(x,\lambda )\)
= \(\overset { \infty }{ \underset { x=0 }{ \Sigma } } x.\frac { e^{ -\lambda }.\lambda ^{ x } }{ x! } \)
Taking out λ from the numerator and x from the denominator

= \(\lambda .{ e }^{ -\lambda }\overset { \infty }{ \underset { x=1 }{ \Sigma } } \frac { { \lambda }^{ x-1 } }{ (x-1)! } \)
= \(\lambda .e^{ -\lambda }(1+\lambda +\frac { { \lambda }^{ 2 } }{ 2! } +...)\)
= \(\lambda .e^{ -\lambda }.e^{ \lambda }\)   [∵ e^x =1+x+\(\frac { x^{ 2 } }{ 2! } \)+..]
= λ.e⁰ =λ(1) =λ
∴ E(X) = λ
E(X²) = \(\overset { \infty }{ \underset { x=0 }{ \Sigma } } p(x,\lambda )\)
= \(\overset { \infty }{ \underset { x=0 }{ \Sigma } } { x }^{ 2 }.\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
=\(\overset { \infty }{ \underset { x=0 }{ \Sigma } } (x(x-1)+x)\frac { e^{ -\lambda }{ \lambda }^{ x } }{ x! } \)
[∵ x(x-1)+x = x²-x+x = x²]
=\(\overset { \infty }{ \underset { x=0 }{ \Sigma } } (x(x-1).\frac { e^{ -\lambda }{ \lambda }^{ x } }{ x! } +\overset { \infty }{ \underset { x=0 }{ \Sigma } } \frac { x.e^{ -\lambda }{ \lambda }^{ x } }{ x! } \)
Taking λ² from the numerator and x(x-1) from. the denominator we get,
E(X²)=\(\overset { \infty }{ \underset { x=0 }{ \Sigma } } \frac { { \lambda }^{ 2 } }{ x(x-1) } .(x)(x-1).\frac { { e }^{ -\lambda }.{ \lambda }^{ x-2 } }{ (x-2)! } +E(X)\)
= \({ \lambda }^{ 2 }.e^{ -\lambda }\overset { \infty }{ \underset { x=0 }{ \Sigma } } \frac { { \lambda }^{ x-2 } }{ (x-2)! } +\lambda \)   [∵ E(X)=λ]
= λ².e^-λ.e^λ+λ   [∵ e^λ =1+λ+\(\frac { { \lambda }^{ 2 } }{ 2! } \)+....]
= λ²+λ
V(X) = E(X²)-[E(X)]² = λ²+ λ - λ² = λ
∴ Mean = Variance = λ

Let p be the probability of success and q be the probability of failure.
Given p = 2q
p = 2(1-p)    [∵ p +  q = 1]
p = 2-2p ⇒ 3p = 2
p = \(\frac{2}{3}\)
∴ q = 1-p = 1-\(\frac{2}{3}\) = \(\frac{1}{3}\) and n = 5
(i) P (3 successes) = P(X = 3)
= ⁵C₃ \(\left( \frac { 2 }{ 3 } \right) ^{ 3 }\left( \frac { 1 }{ 3 } \right) ^{ 2 }\)      ∵ p(x) = ⁿC_x p^xq^n-x, n = 5, x = 3

=\(\frac { 10\times 8 }{ 243 } =\frac { 80 }{ 243 } \)
(ii) P (atleast 3 successes)
= P(X ≥ 3)
= 1-P(X < 3)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= 1-[⁵C₀\(\left( \frac { 2 }{ 3 } \right) ^{ 0 }\left( \frac { 1 }{ 3 } \right) ^{ 5 }\) + ⁵C₁\(\left( \frac { 2 }{ 3 } \right) ^{ 1 }\left( \frac { 1 }{ 3 } \right) ^{ 4 }\) + ⁵C₂\(\left( \frac { 2 }{ 3 } \right) ^{ 2 }\left( \frac { 1 }{ 3 } \right) ^{ 3 }\)]
= 1-\(\left[ \left( \frac { 1 }{ 3 } \right) ^{ 5 }+5\left( \frac { 2 }{ 3 } \right) \left( \frac { 1 }{ { 3 }^{ 4 } } \right) +10\left( \frac { { 2 }^{ 2 } }{ { 3 }^{ 2 } } \right) \left( \frac { 1 }{ { 3 }^{ 3 } } \right) \right] \)
= 1-\(\frac { 1 }{ { 3 }^{ 5 } } \)(1+10+40) = \(1-\frac { 1 }{ 243 } \)(51)
= \(\frac { 243-51 }{ 243 } =\frac { 192 }{ 243 } \).

Let p be the probability of having a laptop
Given p = \(\frac { 40 }{ 100 } \) = 0.4 ⇒ q = 1-p = 1-0.4 = 0.6
n = 15
(i) P(X = 3)
=¹⁵C₃ (0.4)³ (0.6)¹²  ∵ p(x) =ⁿC_x p^xq^n-x, n = 15, x = 3

= 455 (0.064) (0.002176) = 0.0634
(ii) P (12 travellers not having laptop)
= P (3 travellers having laptop)
[∵ n = 15 and 15 - 12 = 3]
= P(X = 3)
= 0.0634.
(iii) P (atleast 3 travellers have laptop)
= P (X ≥ 3) = 1 - P(X < 3)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
=1-[¹⁵C₀ (0.4)⁰ (0.6)¹⁵ + ¹⁵C₁ (0.4)² (0.6)¹³ (0.6)¹⁴]

= 1 - (0.6)¹³ [(0.6)² + 6(0.6)¹ + 105(0.16)]
= 1 - (0.6)¹³ [0.36 + 3.6 + 16.8]
= 1 - (0.0013) (20.76) = 1 - 0.0270
= 0.9730.

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.

Let the probability of student having reading habit
p = 40% = \(\frac { 40 }{ 100 } \) = 0.4
⇒ q = 1-p = 1-0.4 = 0.6
n = 9
(i) P (none of those who have selected having reading habit)
= P(X = 0)
= ⁹C₀(0.4)⁰ (0.6)⁹
= (1)(1)(0.6)⁹   [∵ ⁿC_x p^xq^n-x = p(x), n = 9, x = 0]
= (0.6)⁹   [ ∵ ⁹C₀ = 1 ]
= 0.01008
(ii) P (all those who have selected have newspaper reading habit)
= P(X = 9)
= ⁹C₉(0.4)⁹ (0.6)⁰   [∵ p(x) =ⁿC_x p^xq^n-x, n = 9, x = 9]
= (0.4)⁹   [ ∵ ⁹C₉ = 1 ] 
= 0.000261
(iii) Two thirds of 9 = \(\frac{2}{3}\) x 9 = 6
∴ P(atleast two third have newspaper reading habit)
= P (atleast 6 have newspaper reading habit)
= P(X≥6)
= P(X = 6)+P(X = 7)+P(X = 8)+P(X = 9)
= ⁹C₆ (0.4)⁶ (0.6)³ + ⁹C₇ (0.4)⁷ (0.6)² + ⁹C₈ (0.4)⁸ (0.6)¹ + ⁹C₉ (0.4)⁹ (0.6)⁰
= (0.4)⁶ [⁹C₆ (0.6)³ + ⁹C₇ (0.4) (0.6)² + ⁹C₈ (0.4)² (0.6) + (0.4)³ ]
= (0.4)⁶ [⁹C₃ (0.216) + ⁹C₂ (0.144) + ⁹C₁ (0.096)+0.64     [∵ ⁿC_r = ⁿC_n-r ]
= (0.4)⁶ \(\left[ \frac { 9\times 8\times 7 }{ 3\times 2\times 1 } (0.216)+\frac { 9\times 8 }{ 2\times 1 } (0.144)+9(0.096)+.064 \right] \)
= (0.4)⁶ [18.144 + 5.184 + 0.864 + 0.064]
= (0.4)⁶ [24.256] = (0.0041) (24.256)
∴ P(X≥6) = 0.0994

Given that probability of getting defective item
p = 5% = \(\frac { 5 }{ 100 } \) ⇒ q = 1-p = \(1-\frac { 5 }{ 100 } =\frac { 95 }{ 100 } \)
n = 20
p(x) = \({ n }_{ C_{ x } }{ p }^{ x }{ q }^{ n-x }\), x = 0,1,2....n
(i) P(Exactly 3 defectives)
= \({ 20 }_{ { C }_{ 3 } }\left( \frac { 5 }{ 100 } \right) ^{ 3 }\left( \frac { 95 }{ 100 } \right) ^{ 20-3 }\)
= \(\\ { 20 }_{ { C }_{ 3 } }\)(0.05)3(0.95)17
= \(\\ \frac { 20\times 19\times 18 }{ 3\times 2\times 1 } \) (0.05)³(0.95)⁵(0.95)⁵(0.95)⁵(0.95)²
= (60 x 19) (0.000125)(0.7738)(0.7738)(0.7738)(0.9025)
= 0.059
(ii) P(atleast 2 defectives)
= P(X≥2) =1-P(X<2)
= 1-[P(X=0) + P(X=1)]
= 1-[\(\\ { 20 }_{ { C }_{ 0 } }\)(0.05)⁰(0.95)²⁰ + \(\\ { 20 }_{ { C }_{ 1 } }\)(0.05)¹(0.95)¹⁹]
= 1 - [(0.95)²⁰ + 20 (0.05) (0.95)¹⁹]
= 1 - [0.3585 + (0.3774)]
= 1 - [0.7359] = 0.2641.
(iii) P (exactly 4 defectives)
P(X = 4) = \(\\ { 20 }_{ { C }_{ 4 } }\) (0.05)4(0.95)16

= (15 x 17 x 19) (0.00000625) (0.4402)
= 0.0133
(iv) Find the mean and variance
Mean = np =\(20\times \frac { 5 }{ 100 } =\frac { 100 }{ 100 } \) = 1
Variance =npq = \(20\times \frac { 5 }{ 100 } =\frac { 95 }{ 100 } \) = 0.95

The mean of the binomial distribution
Ex = \(\overset { n }{ \underset { x=0 }{ \Sigma } } x.p(x)=\overset { n }{ \underset { x=0 }{ \Sigma } } \left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
= \(p.\overset { n }{ \underset { x=0 }{ \Sigma } } x.\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x-1 }q^{ n-x }\)
[Take p common]

=  \(np.\overset { n }{ \underset { x=1 }{ \Sigma } } \left( \begin{matrix} n-1 \\ x-1 \end{matrix} \right) { p }^{ x-1 }q^{ n-x }\)
= np (q +p)^n-1 [using binomial theorem]
(x+a)ⁿ = \({ x }^{ n }+{ n }_{ { C }_{ 1 } }{ x }^{ n-1 }{ a }^{ 1 }+...+{ a }^{ n }\)
= np(1)^n-1 [∵ p+q=1]
= np
∴ Mean = E(x) = np..(1)
Now, E(X²) =\(\overset { n }{ \underset { x=0 }{ \Sigma } } { x }^{ 2 }.\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
=\(\overset { n }{ \underset { x=0 }{ \Sigma } } \{ x(x-1)+x\} \left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
=\(\overset { n }{ \underset { x=0 }{ \Sigma } } x(x-1).\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }+{ \Sigma }_{ x }.\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
=\(\overset { n }{ \underset { x=2 }{ \Sigma } } x(x-1)\frac { n(n-1) }{ x(x-1) } \left( \begin{matrix} n-2 \\ n-2 \end{matrix} \right) { p }^{ x-2 }{ q }^{ n-x }\)+\(\Sigma x\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }q^{ n-x }\)
= \(n(n-1){ p }^{ 2 }\left\{ \Sigma \left( \begin{matrix} n-2 \\ n-2 \end{matrix} \right) { p }^{ x-2 }{ q }^{ n-x } \right\} \) +np (using (1))
= n(n-1)p² (q+p)^n-2 + np
[using binomial theory]
= n(n-1)p² (1) + np  ∴ p+ q = 1
= n(n-1)p²+ np .... (2)
Variance = E(X²) - [E(X)]²
= n(n-1)p²+np-(np)²
[From (1) & (2)]

= np (1-p) = npq       [ ∵ p+q = 1⇒ q = 1-p]
∴ Mean = np and variance = npq

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 waiting time of a customer in the queue and it is normally distributed with mean 5 and SD 0.7.

(i) for less than 6 minutes
\(Z=\frac { X-\mu }{ \sigma } =\frac { 6-5 }{ 0.7 } =1.4285\)
P(X < 6) = P(Z < 1.43)
= 0.5 + 0.4236
= 0.9236
(ii) between 3.5 and 6.5 minutes
When X = 3.5
\(Z=\frac { X-\mu }{ \sigma } =\frac { 3.5-5 }{ 0.7 } =2.1429\)
When X = 6.5
\(Z=\frac { X-\mu }{ \sigma } =\frac { 6.5-5 }{ 0.7 } =2.1429\)
P(3.5 < X < 6.5)
= P (–2.1429 < Z < 2.1429)
= P(0 < Z < 2.1429) + P(0 < Z < 2.1429)
= 2 P(0 < Z < 2.1429)
= 2 x .4838
= 0.9676

Let X be the normal variate of life of light bulbs with mean 125 and standard deviation 18.
(i) less than 95 days
When X = 95
\(Z=\frac{X-\mu}{\sigma}=\frac{95-125}{18}=-1.667\)
\(P(X<95)=P(Z<-1.667)\)
= 0.5 – P(0 < Z < 1.67)
= 0.5 – 0.4525
= 0.0475
No. of bulbs expected to expire in less than 95 days out of 900 bulbs 900 × .0475 = 43 bulbs

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

Given that mean μ = 150 and standard deviation σ = 15
(i) when X = 125 thousand
\(Z=\frac { X-\mu }{ \sigma } =\frac { 125-150 }{ 15 } =-1.667\)
When X = 145 thousand
\(Z=\frac { X-\mu }{ \sigma } =\frac { 145-150 }{ 15 } =-0.33\)
Area between Z = 0 and Z = -1.67 is 0.4525
Area between Z = 0 and Z = -0.33 is 0.1293
P(–1.667 ≤ Z ≤ –0.33) = 0.4525 – 0.1293
= 0.3232
Therefore the number of branches having sales between 125 thousand and 145 thousand is 550 × 0.3232 = 178

(ii) When X = 140 thousand
\(Z=\frac { X-\mu }{ \sigma } =\frac { 140-150 }{ 15 } =-0.67\)
When X = 160 thousand
\(Z=\frac { X-\mu }{ \sigma } =\frac { 160-150 }{ 15 } =0.67\)
P(–0.67 < Z < 0.67) = P(–0.67 < Z < 0) + P(0 < Z < 0.67)
= P(0 < Z < 0.67) + P(0 < Z < 0.67)
= 2 P(0 < Z < < 0.67)
= 2 × 0.2486
= 0.4972
Therefore, the number of branches having sales between Rs. 140 thousand and Rs. 160 thousand = 550 × 0.4972 = 273

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

(i) greater than 1.09
The total area under the curve is equal to 1 , so that the total area to the right Z = 0 is 0.5 (since the curve is symmetrical). The area between Z = 0 and 1.09 (from tables) is 0.3621
P(Z > 1.09) = 0.5000 - 0.3621 = 0.1379
The shaded area to the right of Z = 1.09 is the probability that Z will be greater than 1.09

(ii) less than –1.65
The area between -1.65 and 0 is the same as area between 0 and 1.65. In the table the area between zero and 1.65 is 0.4505 (from the table). Since the area to the left of zero is 0.5 , P(Z< 1.65) = 0.5000 – 0.4505 = 0.0495.

(iii) lying between -1.00 and 1.96
The probability that the random variable Z in between -1.00 and 1.96 is found by adding the corresponding areas :
Area between -1.00 and 1.96 = area between (-1.00 and 0) + area betwn (0 and 1.96)
P(–1.00 < Z < 1.96) = P(–1.00 < Z < 0) + P(0 < Z < 1.96)
= 0.3413 + 0.4750 (by tables)
= 0.8163

(iv) lying between 1.25 and 2.75
Area between Z = 1.25 and 2.75 = area betwn (z = 0 and z = 2.75)
– area betwn (z=0 and z = 1.25)
P(1.25 < Z < 2.75) = P(0 < Z < 2.75) – P(0 < Z < 1.25)
= 0.4970 - 0.3944 = 0.1026

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 = 1/5/100 = 1/500 = 0.002 n = 10  λ = np = 0.02
\(p(x)=\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } =\frac { { e }^{ -0.02 }{ (0.02) }^{ x } }{ x! } \)
(i) Number of packets containing no defective = N p(o) = 1,00,000 × e^–0.02
= 98020
(ii) Number of packets containing one defective = N p(1) = 1,00,000 × 0.9802 × 0.02
= 1960
(iii) Number of packets containing 2 defectives = N p(2) = 20

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

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 \({ 32C }_{ x }{ \left( \frac { 1 }{ 2 } \right) }^{ x }{ \left( \frac { 1 }{ 2 } \right) }^{ 32-x }\)

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