11Dmean


 * Binomial Distribution: **

Expected Value, Variance and Standard Deviation

Recall that for __**any**__ probability distribution: **Mean:** math . \qquad \mu = E(X) = \sum \, x \, p(x) \qquad. math

**Variance:** math . \qquad \sigma^2 = Var(X) = E(X^2) - \mu^2 \qquad \{ \textit{where } E(X^2) = \sum \, x^2 \, p(x) \} \qquad. math

**Standard Deviation:** math . \qquad \sigma = SD(X) = \sqrt{Var(X)} \qquad. math

We could use these rules for a binomial distribution, but there are shortcuts.

For any Binomial Distribution:

... ... ** X ~ Bi(n,p) **

where
 * n = number of trials and
 * p = probability of success and
 * q = probability of fail (q = 1 – p)

We can use the following (simpler) rules:

math \\ . \qquad \textbf{Mean: } \qquad \mu = np \\. \\ . \qquad \textbf{Variance: } \: \sigma^2 = npq \qquad. \\ . \\ . \qquad \textbf{S.D. : } \qquad \sigma = \sqrt{npq} math


 * Example 1 **

A fair die is rolled 15 times. Let X = the number of 3s that are rolled. So X ~ Bi(15, 1/6)

a) Find the expected number of 3s rolled

math \\ . \qquad \mu = np \\. \\ . \qquad \;\; = 15 \times \frac{1}{6} \qquad. \\ . \\ . \qquad \;\;= 2.5 math

b) Find the Standard Deviation

math \\ . \qquad \sigma = \sqrt{npq} \\. \\ . \qquad \;\; = \sqrt{15 \times \frac{1}{6} \times \frac{5}{6} } \qquad. \\ . \\ . \qquad \;\; = 1.44 math

c) Find the probability that the number of 3s lies within 2 standard deviations of the mean.

math \\ . \qquad \mu - 2\sigma = 2.5 - 2 \times 1.44 = -0.38 \qquad .\\. \\ . \qquad \mu + 2\sigma = 2.5 + 2 \times 1.44 = 5.38 math

math \\ . \qquad Pr \big( \mu-2\sigma \leqslant X \leqslant \mu + 2\sigma \big) \qquad. \\ . \\ . \qquad \qquad = Pr \big( -0.38 \leqslant X \leqslant 5.38 \big) \\. \\ . \qquad \qquad = Pr \big( 0 \leqslant X \leqslant 5 \big) \\. \\ . \qquad \qquad \textbf{Enter: } binomialcdf \;\; lwr = 0, \; upr = 5, \; n = 15, \; p = 1/6 \qquad. \\ . \\ . \qquad \qquad = 0.9726 \quad \text{ or } 97.3 \% math
 * so **

{This is close to the anticipated result of 95%}

NOTE: Remember that BinomialPDF and BinomialCDF are calculator notations. Don't give calculator notations as a part of your written working on SACs and exams.


 * Example 2 **

The probability of winning a prize in a game of chance is 0.48. ... ... a) What is the least number of games that must be played to ensure that the probability of winning at least once is more than 0.95? ... ... b) What is the least number of games that must be played to ensure that the probability of winning at least twice is more than 0.95?

... ... a) What is the least number of games that must be played to ensure that the probability of winning at least once is more than 0.95?
 * Solution:**

math . \qquad \text{Required to find n, where: } \; X \sim Bi(n, 0.48) \; \text{ and } \; Pr(X \geqslant 1) > 0.95 \qquad. math

math . \qquad \text{Pr}(X \geqslant 1) > 0.95 \\. \\ . \qquad 1 - \text{Pr}(X = 0) > 0.95 \qquad. \\.\\ . \qquad \text{Pr}(X = 0) < 0.05 math

math . \qquad \text{Pr}(X=x) = ^nC_x \; p^x \; q^{n-x} \qquad. \\ . \\ . \\ . \qquad ^nC_0 \; (0.48)^0 \; (0.52)^n < 0.05 \qquad. \\ . \\ . \qquad (0.52)^n < 0.05 \\. \\ math

... ... ... From here, n could be found via trial-and-error. ... ... ... Or we could use logs:

math . \qquad n > \log_{0.52}(0.05) \\ .\\ . \qquad \quad > \dfrac{\log_{10}(0.05)}{\log_{10}(0.52) } \\. \\ . \qquad \quad > 4.6 \qquad \textit{round up to next integer} \qquad. \\ . \\ . \qquad n = 5 math

Hence, min number of games required is n = 5

... ... b) What is the least number of games that must be played to ensure that the probability of winning at least twice is more than 0.95?

math . \qquad \text{Required to find n, where: } \; X \sim Bi(n, 0.48) \; \text{ and } \; Pr(X \geqslant 2) > 0.95 \qquad. math

math . \qquad \text{Pr}(X \geqslant 2) > 0.95 \\. \\ . \qquad 1 - \text{Pr}(X = 0) - \text{Pr}(X=1) > 0.95 \qquad. \\.\\ . \qquad \text{Pr}(X = 0) + \text{Pr}(X=1) < 0.05 math

math . \qquad ^nC_0 \; (0.48)^0 \; (0.52)^n + ^nC_1 \; (0.48)^1 \; (0.52)^{n-1} < 0.05 \qquad. \\ . \\ . \qquad (0.52)^n + n(0.48)(0.52)^{n-1} < 0.05 math

... ... ... From here, find n using trial and error. ... ... ... This can be done on the CAS by typing in the LHS followed by | n = ? ... ... ... And trying different values for n

math . \qquad \text{Hence, min number of games is } \; n = 8 \qquad. math

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