Question

You may need to use the appropriate appendix table to answer this 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 ,

(a)

P(x $\geq$ $90000) = 1 - P(x  $\leq$ 90000)

= 1 - P[(x - $\mu$ ) / $\sigma$ $\leq$ (90000 - 75847) / 33800]

= 1 -  P(z $\leq$ 0.42)   

= 1 - 0.6628

= 0.3372

Probability = 0.3372

(b)

P(x $\leq$ $30000)

= P[(x - $\mu$ ) / $\sigma$ $\leq$ (30000 - 75847) / 33800]

= P(z $\leq$ -1.36)

= 0.0869

Probability = 0.0869

(c)

P($60000 < x < 70000) = P[(60000 - 75847)/ 33800) < (x - $\mu$ ) /$\sigma$  < (70000 - 75847) / 33800) ]

= P(-0.47 < z < -0.17)

= P(z < -0.17) - P(z < -0.47)

= 0.4325 - 0.3192

= 0.1133

Probability = 0.1133

(d)

Using standard normal table ,

P(Z < z) = 86%

P(Z < 1.08) = 0.86

z = 1.08

Using z-score formula,

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

x = 1.08 * 33800 + 75847 = 112351

$112351