11 3meansd

# Binomial Distribution:

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## Expected Value, Variance and Standard Deviation

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Recall that for any discrete probability distribution:
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… … Mean: … … $\mu = E(X) = \sum \, x \, p(x)$
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… … Variance:$\sigma^2 = Var(X) = E(X^2) - \mu^2$
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… … … … … … … where $E(X^2) = \sum \, x^2 \, p(x)$
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… … Standard Deviation:$\sigma = SD(X) = \sqrt{Var(X)}$
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We could use these rules for a binomial distribution, but there are shortcuts.

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For any Binomial Distribution:$X \sim Bi \big(n, \;p \big)$

… … where

… … … n = number of trials

… … … p = probability of success

… … … q = probability of fail $\big( q = 1 - p \big)$
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We can use the following (simpler) rules:

… … Mean: … … $\mu = np$
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… … Variance:$\sigma^2 = npq$
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… … SD: … … . $\sigma = \sqrt{npq}$

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### Example 1

A fair die is rolled 15 times.

Let X = the number of 3s that are rolled.

So $X \sim Bi \Big(15, \; \dfrac{1}{6} \Big)$

a) .. Find the expected number of 3s rolled

b) .. Find the Standard Deviation

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

a) .. Find the expected number of 3s rolled
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… … … … $\mu = np$
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… … … … … $= 15 \times \dfrac{1}{6}$
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… … … … … $= 2.5$

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b) .. Find the Standard Deviation
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… … … … $\sigma = \sqrt{npq}$
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… … … … … $= \sqrt{15 \times \dfrac{1}{6} \times \dfrac{5}{6} }$
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… … … … … $= 1.44$

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c) .. Find the probability that the number of 3s lies within 2 standard deviations of the mean.
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… … … … $\mu - 2\sigma = 2.5 - 2 \times 1.44 = -0.38$
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… … … … $\mu + 2\sigma = 2.5 + 2 \times 1.44 = 5.38$
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… … … so
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… … … … $Pr \big( \mu-2\sigma \leqslant X \leqslant \mu + 2\sigma \big)$
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… … … … … $= Pr \big( -0.38 \leqslant X \leqslant 5.38 \big)$
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… … … … … $= Pr \big( 0 \leqslant X \leqslant 5 \big)$
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… … … … Use calc: binomialcdf lwr = 0, upr = 5, n = 15, p = 1/6
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… … … … … $= 0.9726 \quad \text{ or } 97.3 \%$

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

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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.

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### 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?
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Solution:

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?
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… … … Required to find n, where: $X \sim Bi \big(n, \; 0.48\big)$

… … … and $Pr(X \geqslant 1) > 0.95$

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… … … $Pr(X \geqslant 1) > 0.95$
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… … … $1 - Pr(X = 0) > 0.95$
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… … … $Pr(X = 0) < 0.05$
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… … … $Pr(X=x) = ^nC_x \; p^x \; q^{n-x}$
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… … … $^nC_0 \; (0.48)^0 \; (0.52)^n < 0.05$
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… … … $1 \times 1 \times (0.52)^n < 0.05$
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… … … $(0.52)^n < 0.05$
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… … … From here, n could be found via trial-and-error.

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… … … $n = 5$ is first value of n that gives $Pr < 0.05$
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… … … Or we could use logs:
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… … … $(0.52)^n < 0.05$
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… … … $n > \log_{0.52}(0.05)$
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… … … $n > \dfrac{\log_{10}(0.05)}{\log_{10}(0.52) }$
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… … … $n > 4.6$ … round up to next integer
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… … … $n = 5$
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… … … Hence, min number of games required is $n = 5$

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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?
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… … … Required to find n, where: $X \sim Bi \big(n, \; 0.48\big)$

… … … and $Pr(X \geqslant 2) > 0.95$
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… … … $Pr(X \geqslant 2) > 0.95$
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… … … $1 - Pr(X = 0) - Pr(X=1) > 0.95$
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… … … $Pr(X = 0) + Pr(X=1) < 0.05$
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… … … $^nC_0 \; (0.48)^0 \; (0.52)^n + ^nC_1 \; (0.48)^1 \; (0.52)^{n-1} < 0.05$
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… … … $(0.52)^n + n(0.48)(0.52)^{n-1} < 0.05$
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… … … From here, find n using trial and error.
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… … … This can be done on the CAS by typing in the LHS followed by | n = ?
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… … … And trying different values for n

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… … … $n = 8$ is first value of n that gives $Pr < 0.05$
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… … … Hence, min number of games required is $n = 8$

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