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The probability distribution is:

Mean = P = 0.2.

Variance = P(1 – P) 

= 0.2 x 0.8 

= 0.16.

Mean = μ = 100, 

σ = 5.

The Mean = 0.

Range = (0, ∞), 

Mean = n, Variance = 2n.

(a) The curve is symmetrical β₁ = 0. 

(B)The curve is Mesokurtic β₃ = 3.

λ is parameter; 

Mean = λ; var (x) 

= λ, Here Mean 

= Var (x);

Mean = E(x) = np 

= 6 × \(\frac{1}{3}\) = 2.

P = \(\frac{1}{2}\) = 0.5 B.d is symmetrical.

Mean = np 

= 2, 8p = 2; p 

= \(\frac{2}{8}\) = 0.25 

ie. P = 0.25

For a Binomial Variate Mean > Variance;but here Mean(4) < Variance (5), statement is incorrect.

n = 5, Mean = np = 3 

Put n = 5 in np = 3; 5p 

= 3; p = \(\frac{3}{5}\) = 0.6; 

P = 0.6; Var(X) = npq 

= 5 x 0.6(1 – 0.6) 

= 5 x 0. 6 x 0.4 = 1.2.

The p.m.f of Binomial distribution is:

p(x) = n c_x p^x q^n-x ; x = 0,1,2, ……n.

Here n = 5, p = 0.3 ; 

∴ q = 1 – p = 0.7.

∴ p(x) = 5c_x 0.3^x 0. 7^5-x ; 

x = 0, 1, 2, ….. 5.

Variance = \(\frac{n}{n-2}\) = \(\frac{6}{6-2}\) 

= 1.5.

2(Area from 0 to 1.7) 

= 2(0.4554) 

= 0.9108.

From the table λ = 3.

Total area under the normal curved is one i.e. 1.

S.D.= √var = √npq = √2;

squaring both sides; 

npq = 2; 4q = 2;

∴ q = 0 

∴ 5, p 

= 0.5 & put p 

= 0.5 in np = 4; 

∴ n = 8

No, mean is greater than variance.

SD = √(var(x)) = √2n 

= √(2 x 8) = 4.

Normal distribution.

In poisson distribution mean = variance 

∴ variance = 4.

Here z~N(0, 1) 

∴ p[-1 .64 < Z < 1.64] 

= Area from 0 to (-1.64) + area from 0 to 1.64 

∴ From the table = 0.4495 + 0.4495 

= 0.899.

Q.D = \(\frac{2}{3}\) , x = \(\frac{2}{3}\) x 3  = 2.

Given : n(d.f) = 3; variance 

= 2n = 2 × 3 

= 6.

It is a χ² – distribution with 3 d.f.

(a) n is large (n → ∞) 

(b) neither p nor q is very small.

It is SNV~N(0,1), ie Normal distribution, and χ² – distⁿ with 1 d.f.

Mean = p, Variance = P(1 – P).

The Bernoulli distribution is:

P(x) = PX( 1 – P)^{1 – X} . where P> O, and X = 0,1;

OR p(x) = Pxq^{1 – X} ; x = o,1 (Range); .

P-parameter, Mean = p, Variance = P(1 – p) 0r pq; mean> variance

Ex:- (i) A coin is tossed & getting H or T

(ii) A new born baby is a male or female ¡s a Bernoulli variate with the parameter P = 0.5.

Here Z₁² + Z₂² + Z₃² is a chi-square distribution with 3 degrees of freedom, 

i.e., χ(3)² Variance = 2n = 2.3 = 6.

Area from K to ∞ = 0.10

Area from 0 to k = 0.5 – 0.10 

= 0.4

∴ From the table k = 1.28

1. BD has parameters, 

2. Mean = np,/variance = npq, Here Mean>variance, 

3. BD.is symmetric when p=q.

(a) When n is large i.e., n → ∞ 

(b) Neither p nor q are very small and mean = np = µ , S.D = √(npq) 

= ‘σ’ are the parameters of Normal distribution.

Z² ~ χ² (1) i.e., It is a chi-square distribution with 1 degree of freedom.

Area from 0 to 2.5 – Area from 0 to 1. 

= 0.4938 – 0.3413 

= 0.1525.

Mean = p = \(\frac{1}{3}\) and variance = p(1-p) = \(\frac{1}{3}\)(1 – \(\frac{1}{3}\)) = 0.2211.

SD(x) = √(var(x)) 

= √p(1-p).

Given S.D 

= √variance = 3 

Squaring on both sides

variance = 9

In Poisson distribution mean = variance 

∴ mean = 9.

For the normal distribution

Quartile deviation 

= \(\frac{2}{3}\) σ =  \(\frac{2}{3}\) x 2 

= 1.33.

Mean deviation = \(\frac{4}{5}\) 

= \(\frac{4}{5}\) x 2 = 1.6.

Variance = σ² = 9 

∴σ² = √9 = 3.

In a normal distribution quartile deviation 

= \(\frac{2}{3}σ = \frac{2}{3} \times 3 = 2\).

(a) The total are under the normal curve is unity. 

(b) p [µ – σ < x< µ + σ] 

= 0.6826. 

(c) p [ µ – 2σ < x < µ + 2σ] 

= 0.9544. 

(d) p [µ – 3σ < x < µ + 3σ] 

= 0.9974.

β₁ = 0 and β₂ = 3. 

Mean µ = 0 and variance σ² = 1.

In a Normal distribution the two quartiles are equidistant from the median/ mean / mode.

Mode mode (Z) 

= \(\frac{Q_3+Q_1}{2}\) 

= \(\frac{30+60}{2}\)

= 45.

The S.D. of SNV is σ = 1

A normal variate with mean 

µ = 0 and S.D. σ² 

= 1 is called S.N.V.denoted by Z 

= missing ~N(0,1)

Area from 0 to ∞ = 0.5.

SD = √(Var(x)

2 = √λ 

∴ λ = 4.; 

∴ Mean = 4.

A probability distribution which has the following probability mass function as:

P(x) = (e^-λλ^z/x)

x = 0,1,2,…../λ > 0.

Mean = n, Range is (0, ∞ ) or 0 to ∞ .