1. The mean annual US household income is $72,641. Suppose that the standard deviation is $30,000. a. Find the probability that a household chosen at random has an income greater than $100,000. b. Find the probability that a group of 5 households, chosen at random, have an average income greater than $100,000. c. Find the probability that a group of 10 households, chosen at random, have an average income greater than $100,000.

Answer:

The answer is below

Step-by-step explanation:

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given as:

[tex]z=\frac{x-\mu}{\sigma}\\\\where\ x=raw\ score,\mu=mean\ and\ \sigma=standard\ deviation \\\\For\ a\ sample\ size\ n: \\\\z=\frac{x-\mu}{\sigma/\sqrt{n} }[/tex]

Given that μ = $72,641 and σ = $30,000.

a) x > $100000

[tex]z=\frac{100000-72641}{30000}=0.91[/tex]

P(x > 100000) = P(z > 0.91) = 1 - 0.8186 = 0.1814

b) n = 5

x > $100000

[tex]z=\frac{100000-72641}{30000/\sqrt{5} }=2.04[/tex]

P(x > 100000) = P(z > 2.04) = 1 - 0.9793 = 0.0207

c) n = 10

x > $100000

[tex]z=\frac{100000-72641}{30000/\sqrt{10} }=2.88[/tex]

P(x > 100000) = P(z > 2.88) = 1 - 0.9980 = 0.002

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B: Each runner varies with respect to multiple variables. Each runner should use both models to account for their individual differences. This will provide an accurate comparison for the difference in wear between the models.
Both A and B: Both provide an accurate comparison for the difference in wear between the original model and the new model. The runners are paired in option A based on running ability, and each runner uses both models in option B.
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