Option 4 : Normal

__Explanation:__

**Binomial Distribution:**

A binomial distribution is a common probability distribution that occurs in practice. It arises in the following situation:

- There are n independent trials.
- Each trial results in a "success" or "failure"
- The probability of success in each and every trial is equal to 'p'.

If the random variable X counts the number of successes in the n trials, then X has a binomial distribution with parameters n and p.

X ~ Bin (n, p).

**Properties of Binomial distribution:**

If X ~ Bin (n, p), then the probability mass function of the binomial distribution is

f (x) = P (X =x) = nCr p^{x}(1 - p)^{n - x}

for x = 0, 1, 2, 3,...,n

Mean E (X) = μ = np.

Variance (σ^{2}) = np(1 - p).

**Note:**

- nCr = \(\frac{{n!}}{{x!\left( {n - x} \right)!}}\)
- \( \sum_{x = 0}^{n}\) nCr px(1 - p)n - x = 1

**Theorem:**

Let X_{1}, X_{2}, ..., X_{m} be independent random variables such that X_{i} has a BIn (n_{i}, p) distribution, for i = 1, 2, ..., m. Let

\(Y = \;\mathop \sum \limits_{i = 1}^m {X_i}\)

Then, Y ~ Bin \(\left( {\mathop \sum \limits_{i = 1}^m {n_i},\;p} \right)\)

**Bernoulli Distribution:**

- A Bernoulli experiment/trial has only two possible outcomes, e.g. success/failure, heads/tails, female/male, defective/non-defective, etc.
- The outcomes are typically coded as 0 (failure) or 1 (success).

X ~ Bern (p)

P (X = 1) = 1, P (X = 0) = 1 - p, \(0 ≤ p ≤ 1\)

**Properties:**

- The probability mass function is p(x) = p
^{x}(1 - p)^{1 - x}for x = 0, 1. - The mean is E (X) = μ = (1 × p) + 0 × (1 - p) = p
- Since E (X
^{2}) = (1^{2}× p) + 0^{2}× (1 - p) = p, - σ
^{2}= var (X) = E (X^{2}) - μ^{2}= p - p^{2}= p (1 - p).

**Note:**

The **Bernoulli distribution is a special case of binomial distribution with n = 1.**

__Exponential Distribution:__

The probability density function of the exponential distribution is,

f (x) = λe^{-λx} x ≥ 0

mean = \(\frac{1}{\lambda}\) , variance = \(\frac{1}{\lambda^2}\)

__Normal Distribution:__

The probability density function of normal distribution is given by,

\(f\left( x \right) = \frac{{1\;}}{{σ \sqrt {2\pi } }}{{\rm{e}}^{ - \frac{1}{2}{{\left( {\frac{{x - μ \;}}{σ }} \right)}^2}}}\)

Where, - ∞ < x < ∞

mean = μ

Variance = σ^{2}

Option 1 :

\({7 \over 27}\)

__Concept:__

Binomial distribution

\(P\left( {x = k} \right) = {n_{{C_k}}}{p^k}{q^{n - k}}\)

where

p = Probability of success in one trial

q = Probability of failure in one trial = 1 – p

n = Total number of independent trials

k = Discrete random variable

__Calculation:__

n = 3

__\(P\left( {each\;colour} \right) = \frac{2}{6} = \frac{1}{3}\)__

\(q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}\)

Using binomial distribution

Probability of getting red on the top face at least twice is

P(x ≥ 2) = P(x = 2) + P(x = 3)

\(P\left( {x \ge 2} \right) = {n_{{C_2}}}{p^2}{q^{n - 2}} + {n_{{C_3}}}{p^3}{q^{n - 3}}\)

\(\begin{array}{l} P\left( {x \ge 2} \right) = {3_{{C_2}}}{\left( {\frac{1}{3}} \right)^2}{\left( {\frac{2}{3}} \right)^1} + {3_{{C_3}}}{\left( {\frac{1}{3}} \right)^3}{\left( {\frac{2}{3}} \right)^0}\\ = \frac{6}{{27}} + \frac{1}{{27}} = \frac{7}{{27}} \end{array}\)

Option 1 : np

**Binomial distribution:**

Let p is the probability that an event will happen in a single trail (called the probability of success) and

q = 1 – p is the probability that an event will fail to happen (probability of failure)

The probability that the event will happen exactly r times in n trails (i.e. x successes and n – r failures will occur) is given by the probability function

\(f\left( x \right) = P\left( {X = r} \right) = {n_{{C_r}}}{p^r}{q^{n - r}}\)

where the random variable X denotes the number of successes in n trials and r = 0, 1, 2, … n

For Binomial distribution,

Mean = μ = np

Variance = σ^{2} = npq

Standard deviation = σ = √(npq)

The expected value is sometimes known as the **first moment of a probability distribution**. The expected value is comparable to the mean of a population or sample.

Therefore, the first moment about the origin of the binomial distribution is,

Mean = μ = npA lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is

Option 2 : 0.1937

**Concept:**

The probability that exactly 2 of the chosen items are defective is given as,

By Binomial distribution,

\(P(x=2)\;=\;{}_{}^n{C_x}~{(p)^x}~{q^{(n - x)}}\)

where, p = probability of success, q = probability of failure

**Calculation:**

**Given:**

n = 10, x = 2, p = 0.1, q = 0.9

Therefore, P(exactly 2 of the chosen items are defective) = ^{10}C_{2} × 0.1^{2 }× 0.9^{10 - 2} ⇒ 45 × 0.01 × 0.43 = 0.1937

Option 4 : 2[1 - P(Z ≤ 2)]

**Concept:**

The Standard Normal Distribution:

The standard normal distribution is a normal distribution with a Mean of 0 and a Standard Deviation of 1.

The standard normal distribution is *centered at zero* and the degree to which a given measurement *deviates from the mean is given by standard deviation.*

If X is normally distributed with mean μ and standard deviation σ, then \(Z = \frac{{X - μ }}{σ }\) is standard normally distributed with mean 0 and standard deviation 1.

**Analysis:**

Given: X ∼ N (μ, σ^{2})

μ^{2} = σ^{2} (μ > 0)

P(X < -μ | X < μ) = ?

Now,

Converting distribution of X ∼ N (μ, σ2) into normal distribution Z ∼ N (0, 1)

Where, \(Z=\dfrac{X-μ}{σ}\)

since, σ^{2} = 1 we have, σ = 1

μ2 = σ2 and μ > 0 so μ = 1

When X = -1,

\(Z=\dfrac{-1-1}{1}=-2\)

When X = 1,

\(Z=\dfrac{1-1}{1}=0\)

P(X < -μ|X < μ) = P (X < -1 | X < 1)

= P (Z < -2 | Z < 0)

\(\rm P(Z<-2|Z<0)=\dfrac{P(Z<-2\cap Z<0)}{P(Z<0)}\)

= \(\rm \dfrac{P(Z<-2)}{P(Z<0)}=\dfrac{P(Z>2)}{P(Z<0)}=\dfrac{1-P(Z<2)}{1/2}\)

= 2 [1 - P(Z < 2)]

1. Normal distribution is symmetric about Y axis So P(X< -a) = P(X> a).

2.Total Area of the Normal distribution curve is 1. The half area lies on the left side of the mean and the half area lies on the right side of the mean.

3. For any random variable X: Half area lies on the left side of mean P(X > a) = 1- P(X < a).

__Concept:__

**Binomial distribution**

**\(P\left( {x = k} \right) = {n_{{C_k}}}{p^k}{q^{n - k}}\)**

where

p = Probability of success in one trial

q = Probability of failure in one trial = 1 – p

n = Total number of independent trials

k = Discrete random variable

__Calculation:__

n = 3

__\(P\left( {each\;colour} \right) = \frac{2}{6} = \frac{1}{3}\)__

\(q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}\)

Using binomial distribution

Probability of getting red on the top face at least twice is

P(x ≥ 2) = P(x = 2) + P(x = 3)

\(P\left( {x \ge 2} \right) = {n_{{C_2}}}{p^2}{q^{n - 2}} + {n_{{C_3}}}{p^3}{q^{n - 3}}\)

\(\begin{array}{l} P\left( {x \ge 2} \right) = {3_{{C_2}}}{\left( {\frac{1}{3}} \right)^2}{\left( {\frac{2}{3}} \right)^1} + {3_{{C_3}}}{\left( {\frac{1}{3}} \right)^3}{\left( {\frac{2}{3}} \right)^0}\\ = \frac{6}{{27}} + \frac{1}{{27}} = \frac{7}{{27}}=0.259 \end{array}\)

__ Answer__: 15 to 16

__Data__

p = probability of success event = 0.4

q = probability of failure event = 0.6

n = Number of trials

__ Formulae__:

Standard Deviation = \(\sqrt {Variance\;}\)

Variance in Binomial Distribution: npq

** Calculation**:

Standard Deviation = \(\sqrt {npq\;}\)

= \(\sqrt {1000\times 0.4\times 0.6\;}\)

=15.49 ≈15

Option 3 : 25

__Concept:__

Mean = μ = np

Standard deviation = \( {\rm{\sigma }} = \sqrt {{\rm{npq}}}\)

Where n = number of trials;

p = probability of success;

q = (1 - p) = probability of failure

__Calculation:__

Given: Mean = μ = np = 5

Variance = σ^{2} = npq = 4

⇒ 4 = np (1 − p) = 5(1 − p)

⇒ \(\frac{4}{5}\) = (1 − p)

⇒ p = 1 – \(\frac{4}{5}\)

∴ p = \(\frac{1}{5}\)

Again

μ = np = 5

∴ n = \(\frac{5}{p}\) = 5 × 5 = 25Option 1 : \(\frac{3}{{64}}\)

**Concept:**

**Binomial distribution**

\(P\left( {X = r} \right) = {n_{{c_r}}}{p^r}{q^{n - r}}\)

Mean = np

Variance = npq

Standard deviation \( = \sqrt {npq} \)

__Calculation__:

Mean = np = 1

Variance = npq = 3/4

\( \Rightarrow p = \frac{1}{4},q = \frac{3}{4},n = 4\)

\(P\left( {X = 3} \right) = {4_{{c_3}}}{\left( {\frac{1}{4}} \right)^3}{\left( {\frac{3}{4}} \right)^{4 - 3}} \)

\(P(X=3)= 4 \times \frac{1}{{64}} \times \frac{3}{4} = \frac{3}{{64}}\)

Option 3 : 323

__Explanation:__

**Given: **

Mean number of defectives = 2 = np

⇒ 2 = 20p

__Calculation: __

\({\rm{Hence}},{\rm{\;probability\;of\;defective\;part\;}} = p = \frac{2}{{20}} = 0.1\)

Probability of non-defective = 0.9

**Now,**

∴ Probability of at least three defectives in a sample of 20 = 1 – [P(0) + P(1) + P(2)]

\( = 1 - \left[ {{}_{}^{20}{C_0}{{\left( {0.9} \right)}^{20}} + {}_{}^{20}{C_1}\left( {0.1} \right){{\left( {0.9} \right)}^{19}} + {}_{}^{20}{C_2}{{\left( {0.1} \right)}^2}{{\left( {0.9} \right)}^{18}}} \right]\)

**= 0.323**

Thus, the number of samples having at least three defective parts out of 1000 samples

= 1000 × 0.323

**= 323**