An article in the European Physical Journal B ["Evidence for the exponential distribution of income in the USA" (2001, Vol. 20, pp. 585-589)] shows that individual income in the US can be approximated with an exponential distribution. Suppose that the mean of individual income in a year is $23000. (a) Write the probability distribution function for individual income. f (x) = Edit LINK TO TEXT (b) What percentage of individuals in the US earn less than $21000? Round your answer to three decimal places (e.9. 98.765). P(X < 21000) = L (c) What percentage of individuals in the US earn more than $495007 Round your answer to three decimal places (e. g. 98.765). P(X > 49500) = (d) What is the income exceeded by 2% of individuals?

a) f(x) = P(X=x) = λ e^(-λ.x)

b) P(X < 21000) = 0.59868 = 59.868%

c) P(X > 49500) = 0.11623 = 11.623%

d) $464.66

Step-by-step explanation:

a) Exponential random variable's probability function is given as

f(x) = P(X=x) = λ e^(-λ.x)

The cumulative distribution function is given as

P(X < x) = 1 - e^(-λ.x)

where λ = rate parameter = (1/mean)

b) The percentage of individuals in the US earn less than $21000

P(X < 21000)

P(X < x) = 1 - e^(-λ.x)

x = 21000

λ = (1/23000) = 0.0000434783

λ × x = 0.0000434783 × 21000 = 0.913

P(X < 21000) = 1 - e⁻⁰•⁹¹³ = 0.59868 = 59.868%

c) The percentage of individuals in the US earn more than $49500

P(X > 49500)

P(X > x) = 1 - [1 - e^(-λ.x)] = e^(-λ.x)

x = 49500

λ = (1/23000) = 0.0000434783

λ × x = 0.0000434783 × 49500 = 2.152

P(X > 49500) = e⁻²•¹⁵² = 0.1162312064 = 0.11623 = 11.623%

d) What is the income exceeded by 2% of individuals?

P(X < x) = 2% = 0.02

P(X |< x) = 1 - e^(-λ.x) = 0.02

e^(-λ.x) = 0.98

In [e^(-λ.x)] = In 0.98 = -0.0202027073

-λ.x = -0.0202027073

x = 0.0202027073 ÷ λ

x = 0.0202027073 ÷ 0.0000434783

x = $464.66

Hope this Helps!!!

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