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 $50,000 or less? (Round your answer to four decimal places.)

(c) What is the probability that a household in Maryland has an annual income between $40,000 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

Answer:

a. 0.3372 or 33.72%; b. 0.2236 or 22.36%; c. 0.2879 or 28.79%; d. $112,351.00

Step-by-step explanation:

We need to use here the values from the cumulative standard normal table and z-scores to solve the questions. That is, all the values are transformed to a z-score to use the standard normal table to find the probabilities. Notice that the question is telling us about the median and not mean. Fortunately, in the normal distribution the mean, the median, and the mode are the same. So, we can say that:

$\\ median\;,mode\;,and\;mean=\mu$

As a result, the parameters for the normal distribution in this case are:

$\\ \mu = 75847\;and\;\sigma=33800$

Then we can solve the questions as follows:

Part a: Probability that annual income of 90,000 or more

We need to calculate the z-score of such a value of x=90,000:

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{90000 - 75847}{33800}$

$\\ z = \frac{14153}{33800}$

$\\ z = 0.41872$

We need to round this value to z = 0.42 to use the cumulative standard normal table. This value is above the mean (positive) and corresponds, approximately, with a cumulative probability of P(x<90000) = 0.66276.

Then, the probability that a household in Maryland has an annual income of $90,000 or more is:

$\\ P(x\geq90000) = 1 - P(x<90000) = 0.33724$

Rounding to four decimal places is 0.3372 or 33.72%

We can follow the same procedure to find the rest of the probabilities asked.

Part b: Probability a household has an annual income of 50,000 or less.

$\\ P(x\leq50000) = \?$

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{50000 - 75847}{33800}$

$\\ z = -0.764704$

Rounding this value to two decimals (z = -0.76), we can conclude that this value is below the mean. To find this probability from the cumulative standard normal table, we first found the value of z = 0.76 (since no negative value is displayed in this table) and then subtracting this value from one. This is possible because the normal distribution is symmetrical. Then,

$\\ P(z<0.76) = 0.77637$

$\\ P(z<-0.76) = 1 - P(z<0.76) = 1 - 0.77637 = 0.22363$

Thus, the probability that a household in Maryland has an annual income of $50,000 or less is 0.2236 (rounding to four decimals) or 22.36%.

Part c: Annual income between $40,000 and $70,000

Mathematically, it can be expressed as:

$\\ P(40000<x<70000) = \?$

For x = 40000:

$\\ z = \frac{40000 - 75847}{33800}$

$\\ z = -1.06085$

This value is below the mean and is -1.06085 standard deviations from it.

Following the same procedure in Part b, the value for z = -1.06 corresponds to a cumulative probability of (P(z<1.06) = 0.85543):

$\\ P(z<-1.06) = 1 - 0.85543 = 0.14457$

For x = 70000:

$\\ z = \frac{70000 - 75847}{33800}$

$\\ z = -0.17298$

Which corresponds to a cumulative probability of (P(z<0.17) = 0.56749):

$\\ P(z<-0.17) = 1 - 0.56749 = 0.43251$

Then, the probability that a household in Maryland has an annual income between $40,000 and $70,000 is:

0.43251 - 0.14457 = 0.28794.

Rounding to four decimals is 0.2879 or 28.79%.

Part d: Eighty-sixth percentile

The z-score that corresponds to a probability of 86% or 0.86 is z = 1.08.

Then, solving the equation for the corresponding z-score:

$\\ 1.08 = \frac{x - 75847}{33800}$

$\\ 1.08*33800= x - 75847$

$\\ 1.08*33800 + 75847 = x$

$\\ 1.08*33800 + 75847 = x$

$\\ x = 112351$

Then, the annual income of a household in the eighty-six percentile of annual household in Maryland is $112,351.00.

Answer 2

The annual income (in $) of a household in the eighty-sixth percentile of annual household income in Maryland is $112,351.

What is Probability?

The probability helps us to know the chances of an event occurring.

$\rm Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}$

As it is given that the median income of Maryland is $75,847(μ) and the standard deviation is $33,800(σ). Also, it is given that annual household income in Maryland follows a normal distribution.

A.) In order to find the probability that a household in Maryland has an annual income of $90,000 or more. We will write it as,

$P(X\geq 90,000) &= 1-P(X < 90,000)$

                        $= 1-P(Z < \dfrac{X-\mu}{\sigma}= \dfrac{90,000-75,847}{33,800})\\\\= 1-P(Z < \dfrac{90,000-75,847}{33,800})\\\\= 1 -P(Z < 0.4187)\\\\= 1-0.6628\\\\=0.3372 = 33.72\%$

B.) In order to find the probability that a household in Maryland has an annual income of $50,000 or less. We will write it as,

$P(X\leq 50,000) &= P(X \leq 50,000)$

                        $= P(Z \leq \dfrac{X-\mu}{\sigma}= \dfrac{50,000-75,847}{33,800})\\\\= P(Z \leq \dfrac{50,000-75,847}{33,800})\\\\= P(Z\leq -0.7647)\\\\= 0.2236\\\\=22.36\%$

C.)  In order to find the probability that a household in Maryland has an annual income between $40,000 and $70,000,

$P(40,000 < X < 90,000) = P(X_1 < 90,000) - P(X_2 < 40,000)$    

                        $= P(Z_1 < \dfrac{X_1-\mu}{\sigma}= \dfrac{90,000-75,847}{33,800}) - P(Z_2 < \dfrac{X_2-\mu}{\sigma} = \dfrac{40,000-75,847}{33,800})\\\\= P(Z_1 < 0.42)- P(Z_2 < -1.060)\\\\=0.6628 - 0.1446\\\\=0.5182$

 

D.) The annual income (in $) of a household in the eighty-sixth percentile of annual household income in Maryland. Therefore, we can write it as,

$P(X\leq x) = 86\%$

$P(Z\leq z=\dfrac{x-\mu}{\sigma})=0.86$

Using the Z-table, for Z=1.08, the P-value is approximately 86%.

$z= 1.08 \implies \dfrac{x-\mu}{\sigma} =1.08\\\\x = 1.08\sigma + \mu\\\\x = (1.08 \times 33,800) + 75,847 \\\\x = 112,351$

Hence, the annual income (in $) of a household in the eighty-sixth percentile of annual household income in Maryland is $112,351.

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