Question

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

a. What is the probability that a household in Maryland has an annual income of $100,000 or more?

b. What is the probability that a household in Maryland has an annual income of $40,000 or less?

c. What is the probability that a household in Maryland has an annual income between $50,000 and $70,000?

d. What is the annual income of a household in the 90th percentile of annual household income in Maryland?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 75847

standard deviation = $\sigma$ = 33800

a) P(x $\geq$ 100000) = 1 - P(x  $\leq$ 100000)

= 1 - P[(x - $\mu$ ) / $\sigma$ $\leq$ (100000 - 75847) / 33800 ]

= 1 -  P(z $\leq$ 0.71)   

  Using z table,

= 1 - 0.7611

= 0.2389

b) P(x $\leq$ 40,000)

= P[(x - $\mu$ ) / $\sigma$ $\leq$ (40,000 - 75847) / 33800]

= P(z $\leq$ -1.06)

Using z table,

= 0.1446

c) P(50000 < x < 70000) = P[(50000 - 75847)/ 33800) < (x - $\mu$ ) /$\sigma$  < (70000 - 75847) / 33800) ]

= P(-0.76 < z < -0.17)

= P(z < -0.17) - P(z < -0.76)

Using z table,

= 0.4325 - 0.2236

= 0.2089

d) Using standard normal table,

P(Z < z) = 90%

= P(Z < z) = 0.90  

= P(Z < 1.282) = 0.90

z = 1.282

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 1.282 * 33800 + 75847

x = 119178.6

90 th percentile is = $119,179

Answer 2

the last one should be =1.28*3380+75847

which gives a cleaner answer of = $119,111