Question

According to Money magazine, Maryland had the highest median annual household income of any state in 2018 at $75,847.† Assume that annual household income in Maryland follows a normal distribution with a median of $75,847 and standard deviation of $33,800.

(a)

What is the probability that a household in Maryland has an annual income of $90,000 or more? (Round your answer to four decimal places.)

(b)

What is the probability that a household in Maryland has an annual income of $30,000 or less? (Round your answer to four decimal places.)

(c)

What is the probability that a household in Maryland has an annual income between $60,000 and $70,000? (Round your answer to four decimal places.)

(d)

What is the annual income (in $) of a household in the eighty-sixth percentile of annual household income in Maryland? (Round your answer to the nearest cent.)

Answer 1

Solution:
Given that,
μ =$75847 , σ=$33800
A) P(X>=90000)=1-P(X<90000)
=1- p{[(x- μ)/σ]<[(90000 - 75847)/33800]}
= 1- P(z< 0.42)
=1- 0.6628 ( from Standard Normanl table)
=0.3372

B) solution;
P(X<= 30000)

= p{[(x- μ)/σ]<=[(30000 - 75847)/33800]}
=P(z< -1.36)
=0.08690 ( from Standard Normanl table)

C)
P(60000<Y< 70000)= p{[(60000 - 75847)/33800]<[(Y- μ)/σ]<[(110 - 75847)/33800]}
=P(-0.47<Z< -0.17)
= p(Z< -0.17) - p(Z< -0.47)
= 0.4325 - 0.3192 ( from. Standard Normanl table)
=0.1133
D)
By using standard normal distribution ,we have to find the corresponding z-values as
P( Z< z)= 86%
p( Z< z)= 0.86
From the standard normal table.
P( Z< 1.08 )= 0.86
z= 1.08
Now using the z-score formula, we can find the value of x as follows,
z=(x- μ)/σ
z×σ=( x- μ)
x= μ+( z×σ)
x=75847+(1.08×33800)
=75847+36504
x=112351
86th percentile is $112351