Question

Suppose 10 percent of households earn over 80.000 dollars a year, and 0.25 percent of households earn over 450,000. A random sample of 400 households has been chosen. In this sample, let X be the number of households that earn over 80,000, and let Y be the number of households that earn over 450,000. Use normal and Poisson approximation, whichever is appropriate in either case, to find the simplest estimates you can for the probabilities P(X 48) and P(Y 2).

Answer 1

Answer:

0.1056

0.2642

Step-by-step explanation:

N = 400

P = 10%

n*p = 400*0.1 = 40

n(1-p) = 400 * 0.9 = 360

40 and 360 are greater than 5 so we use the normal distribution here to estimate further.

mean = n*p = 40

sd = $\sqrt{np(1-p)}$

$\sqrt{400*0.1(1-0.1)} \\= \sqrt{40(0.9)} \\= \sqrt{36} \\= 6$

p(X≥48)

= p(X≥47.5)

$=P(Z$

= 0.8944

1 - 0.8944

= 0.1056

FOR Y

N = 400 p = 0.25% = 0.0025

400 * 0.0025 = 1

1 is less than 5 so we use poisson distribution.

using the excel function,

1 - BINOMDIST(1, 400, 0.0025, TRUE)

= 1 - 0.735759074

= 0.2642