MathsTips.com

Maths Help, Free Tutorials And Useful Mathematics Resources

  • Home
  • Algebra
    • Matrices
  • Geometry
  • Trigonometry
  • Calculus
  • Business Maths
  • Arithmetic
  • Statistics
Home » QnA » Mean and Variance of a binomial distribution are 24 and 128

Mean and Variance of a binomial distribution are 24 and 128

If the sum and the product of the mean and variance of a binomial distribution are 24 and 128, respectively, then the probability of one or two successes is :

  1. \frac{{33}}{{{2^{32}}}}
  2. \frac{{33}}{{{2^{29}}}}
  3. \frac{{33}}{{{2^{28}}}}
  4. \frac{{33}}{{{2^{27}}}}

Solution:

Tip for solving this question:

Firstly find the mean and variance of the binomial distribution.

Make an equation according to the question and solve it.

Then use the formula of probability for the binomial distribution.

Step 1 of 3:

The mean of the binomial distribution is:

Mean = np

And the variance of the binomial distribution is:

Variance = npq

Where n is the number of trials

P is the probability of success

q is probability of failure i.e. q = 1- p ……….(1)

Step 2 of 3:

According to the question,

Sum of mean and variance of binomial distribution = 24

i.e. mean + variance = 24

np + npq = 24

np(1 + q) = 24

np = \frac{{24}}{{1 + q}} ……………….(2)

Product of mean and variance of binomial distribution = 128

i.e. (mean)(product) = 128

(np) (npq) = 128

{\left( {np} \right)^2}q = 128

Using Equation (2)

\begin{gathered} \Rightarrow {\left( {\frac{{24}}{{1 + q}}} \right)^2}q = 128 \hfill \\ \Rightarrow \frac{{576q}}{{{{(1 + q)}^2}}} = 128 \hfill \\ \Rightarrow 576q = 128{(1 + q)^2} \hfill \\ \Rightarrow 576q = 128(1 + {q^2} + 2q) \hfill \\ \Rightarrow 576q = 128 + 128{q^2} + 256q \hfill \\ \Rightarrow 128{q^2} + 256q - 576q + 128 = 0 \hfill \\ \Rightarrow 128{q^2} - 320q + 128 = 0 \hfill \\ \Rightarrow 64(2{q^2} - 5q + 2) = 0 \hfill \\ \Rightarrow 2{q^2} - 5q + 2 = 0 \hfill \\ \end{gathered}

On factorisation, we get

\Rightarrow \left( {q - 2} \right)\left( {2q - 1} \right) = 0

\Rightarrow q = 2 \text{ or } q = \frac{1}{2}

When q = 2\Rightarrow p = 1 – q

\Rightarrow p = 1 – 2 = -1 ………..Not considered due to negative

When q = \frac{1}{2}\Rightarrow p = 1 – q

\Rightarrow p = 1 - \frac{1}{2} = \frac{1}{2}

So, we get p = q = \frac{1}{2}

Using equation (2), n = 32

Step 3 of 3:

The probability for binomial distribution is:

p(X = x) = {}^n{C_x}{p^x}{q^{n - x}}

probability of one or two successes = p(X = 1) + p(X = 2)

= {}^{32}{C_1}{p^1}{q^{32 - 1}}+ {}^{32}{C_2}{p^2}{q^{32 - 2}}

= {}^{32}{C_1}\left( {\frac{1}{2}} \right){\left( {\frac{1}{2}} \right)^{31}}+{}^{32}{C_1}{\left( {\frac{1}{2}} \right)^2}{\left( {\frac{1}{2}} \right)^{30}}

= \left( {32 + \frac{{32 \times 31}}{2}} \right)\left( {\frac{1}{{{2^{32}}}}} \right)

= \frac{{33}}{{{2^{28}}}}

Final Answer:

Hence, Option (C) is correct.

« Line with slope greater than 1 passes A(4, 3) and intersects x–y=2 at B
The slope of the tangent to a curve C: y = y(x) at any point (x, y) »


Filed Under: QnA Tagged With: JEE Main 2022, July 25th Shift 1

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Table of Content

  • For theta Belonging to 0, 4Pi the linear equations with no solutions
  • The total number of functions, f: {1, 2, 3, 4} → {1, 2, 3, 4, 5, 6} such that f(1) + f(2) = f(3), is equal to
  • α^2021 + β^2021 + γ^2021 + δ^2021 for x^4 + x^3 + x^2 + x + 1 = 0
  • Number of elements where n belongs to N and Sn Intersection Tn is phi
  • Lim n →∞, sqrt(n2-n-1)+nα+β=0 then 8(α+β)
  • Absolute maximum value of the function (x^2 – 2x + 7)e^(4x^3 – 12x^2 – 180x + 31)
  • General Solution of differential equation (x – y^2)dx +y(5x + y^2)dy
  • Line with slope greater than 1 passes A(4, 3) and intersects x–y=2 at B
  • Mean and Variance of a binomial distribution are 24 and 128
  • The slope of the tangent to a curve C: y = y(x) at any point (x, y)

© MathsTips.com 2013 - 2023. All Rights Reserved.