Question

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 1

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:

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

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

a) x > $100000

$z=\frac{100000-72641}{30000}=0.91$

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

b) n = 5

x > $100000

$z=\frac{100000-72641}{30000/\sqrt{5} }=2.04$

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

c) n = 10

x > $100000

$z=\frac{100000-72641}{30000/\sqrt{10} }=2.88$

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