Unit 4 Would this get credit?

Multiple Choice

i. $$\mu=E\left(G\right)=\frac{1}{.08}\approx12.5\ geodes$$

ii. $$\sigma_G=\frac{\sqrt[]{1-.08}}{.08}\approx11.99\ geodes$$

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Multiple Choice

i. 12.5

ii. 11.99

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Multiple Choice

i. 13

ii. 12

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Multiple Choice

i. $$\mu=E\left(G\right)=\frac{1}{.08}\approx13\ geodes$$

ii. $$\sigma_G=\frac{\sqrt[]{1-.08}}{.08}\approx12\ geodes$$

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Multiple Choice

i. $$\mu=E\left(G\right)=\frac{1}{p}$$

ii. $$\sigma_G=\frac{\sqrt[]{1-p}}{p}$$

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Multiple Choice

i. 0.0677

ii. 0.0623

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Multiple Choice

i. 0.067712

ii. 0.778688

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Multiple Choice

i. geometpdf(0.08, 3) = 0.0677

ii. 1 - geometcdf(0.08, 3) = 0.7787

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Multiple Choice

i. geometpdf(p=0.08, x=3) = .0677

ii. 1 - geometcdf(p-0.08, x=3) = 0.778688

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Multiple Choice

i. $$P\left(Y=3\right)=\left(0.92\right)^2\left(0.08\right)\approx0.067712$$

ii. $$P\left(Y=4\right)=1-P\left(1\ or\ 2\ or\ 3\right)=1-\left(0.08+0.0736+0.067712\right)=0.778688$$

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Multiple Choice

i. $$\frac{1}{.08+.0736+.0677+.0623}\approx3.526$$

ii. The mean of 3.5 is the average number of geodes opened.

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Multiple Choice

i. 1-VAR STATS(L1, L2) = 3.545

ii. The mean of 3.545 is the average.

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Multiple Choice

i. $$\mu=E\left(Y\right)=1\left(.08\right)+2\left(.0736\right)+3\left(.0677\right)+4\left(.0623\right)\approx0.6795\ geodes$$

ii. The mean of .6795 geodes is the average number of geodes opened.

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Multiple Choice

i. $$\mu=E\left(Y\right)=\left(1\right)\left(0.08\right)+...+4\left(0.778688\right)\approx3.545\ geodes$$

ii. 3.5 is the number of red geodes.

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Multiple Choice

i. $$\mu=E\left(Y\right)=1\left(.08\right)+...4\left(.778688\right)\approx3.545\ geodes$$

ii. The mean of 3.545 geodes is the average number of geodes that results from many many trials of opening randomly selected geodes and counting the number opened until either a red geode is found or the 4th geode is opened.

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