Chapter 6 - Normal Probability Distributions - 6-5 The Central Limit Theorem: 20

a) 0.9999 b)0.7939

Work Step by Step

a) By using the Central Limit Theorem, the sample mean has a mean of $\mu$ and standard deviation of $\frac{\sigma}{\sqrt n}$. $z=\frac{value-mean}{standard \ deviation}=\frac{140-182.9}{\frac{40.8}{\sqrt{16}}}=-7.44.$ Using the table, the probability of z being more than -7.44 is equal to 1 minus the probability of z being less than -7.44, which is: 1-0.0001=0.9999. b) By using the Central Limit Theorem, the sample mean has a mean of $\mu$ and standard deviation of $\frac{\sigma}{\sqrt n}$. $z=\frac{value-mean}{standard \ deviation}=\frac{174-182.9}{\frac{40.8}{\sqrt{16}}}=-0.82.$ Using the table, the probability of z being more than -0.82 is equal to 1 minus the probability of z being less than -0.82, which is: 1-0.2061=0.7939.

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