# Let X follow binomial distribution with parameters n = 32 \times 10^6 and p = 2 \times 10^{-6} ...

## Question:

Let X follow binomial distribution with parameters {eq}n = 32 \times 10^6 {/eq} and {eq}p = 2 \times 10^{-6} {/eq}

(a) What is the {eq}P(52 \leq X \leq 76) {/eq}?

(b) Find P(X = 63)?

(c) Use Chebyshev's theorem to find a bound for P(56 < X < 72). Round standard deviation to the nearest whole number.)

## Binomial Theorem:

Binomial Theorem is a part of algebraic expansion between the constraints to the power given. In the binomial theorem, the given algebraic expression is presented with the power of expansion and to be arranged in the binomial form of constraints.

Given Information

{eq}\begin{align*} n = 32 \times {10^6}\\ p = 2 \times {10^{ - 6}} \end{align*} {/eq}

To Evaluate

{eq}\begin{align*} p\left( {52 \le x \le 76} \right)\\ p\left( {x = 63} \right)\\ p\left( {56 < x < 72} \right) \end{align*} {/eq}

By Chebyshev?s Theorem

Evaluation

As prescribe that P between x1 and x2 = P (at most x2) ? P (at most x1-1)

Where X1 = 52 and X2 = 76

By,

Binomial Distribution we have:

N = number of trails = 32000000

P = the probability of a success = 0.000002

Therefore,

P At Most 51 = 0.05520

P at most 76 = 0.9376

Therefore,

Between X1 and X2 = 0.88248

Probability of success will be:

P (n,x) = nCx p^x(1-p)^(n-x)

Where,

N = number of trails = 32000000

P = the probability of a success = 0.000002

X = Number of successes = 63

Therefore the probability formed is

P(63) = 0.04980

Where,

U = Mean = NP = 64

S = SD = 8

Therefore,

Between 56 and 72, 1 is the standard deviation from the calculated mean

So, by

Chebyshev?s Theorem where k = 1

{eq}\begin{align*} 1 - \dfrac{1}{{{k^2}}} = 1 - \dfrac{1}{{{{\left( 1 \right)}^2}}} = 0\\ p\left( {56 < x < 72} \right) > 0 \end{align*} {/eq}