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In an attempt to increase sales, a breakfast cereal company decides to offer a promotion. Each box of cereal will contain a collectible card featuring one NASCAR driver: Kyle Busch; Dale Earnhardt, Jr.; Kasey Kahne; Danica Patrick; or Jimmie Johnson. The company says that each of the 5 cards is equally likely to appear in the 100,000 boxes of cereal that are part of this promotion. You buy 12 boxes and let X = the number of Kyle Busch cards in the sample. Calculate the mean and standard deviation of the sampling distribution of X.

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We know that if XX is the number of successes in a random sample of size nn from a large population with a proportion of successes pp. Then,

The mean\textbf{mean} of the sampling distribution of XX is given by μ=np\mu=np.

The standard deviation\textbf{standard deviation} of the sampling distribution of XX is given by formula σ=np(1p)\sigma=\sqrt{np(1-p)}.

Here, it is given that all the five cards in the box are equally likely to appear in the 100,000 boxes of cereal that are part of this promotion. If we buy 12 boxes and let XX be the number of Kyle Busch cards in the sample.

The size of the sample is given as n=12n=12 and the proportion of success that a Kyle Busch card is selected out of five cards is 15\dfrac{1}{5} means p=0.2p=0.2.

Thus, by applying the formula stated above, we could say that the mean of the sample distribution of XX is

μX=np=12×0.2=2.4\mu_X=np=12\times 0.2=2.4

and the standard deviation of the sample distribution of XX is

σX=np(1p)=2.4(10.2)=2.4×0.8=1.921.39\sigma_X=\sqrt{np(1-p)}=\sqrt{2.4(1-0.2)}=\sqrt{2.4\times 0.8}=\sqrt{1.92}\approx 1.39

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