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Question

Family incomes in two particular neighborhoods follow a normal distribution.

QUESTION: Example: Family incomes in two particular neighborhoods follow a normal distribution. For Neighborhood 1 the mean is $26,500, with a standard deviation of $15,000. For Neighborhood 2, the mean is $30,000 with a standard deviation of $18,000.

A) What proportion of families in Neighborhood 1 have incomes below the mean family income in Neighborhood 2?

B) What income level in Neighborhood 1 represents the top 10% for that neighborhood?

C) What income level for neighborhood 2 represents the lowest 5% of incomes for that neighborhoods?

D) For neighborhood 1, what percentage of families have incomes that fall between the mean incomes of Neighborhood 1 and neighborhood 2?

 

Solution

M1 = 26 500 M2 = 30,000
<br>STD 1 = 15,000 STD 2 = 18,000
<br>
<br>A)P(X <= 30,000)where X is the THE PROPORTION whose fam income for neighborhood 1 is less than 30,000( MEan Income for neighborhood 2).
<br> Calculate the Z score; Z = (30,000-26,500)/STD 1
<br> = .2333
<br> Therefore Prob = .5922 ( from Normal Table for corresponding Z value)
<br>
<br>B)Top 10% means 90% must be below this income figure X.
<br> There Z score is given in this case , from tables Z= 1.28
<br> 1.28 = (X - 26,500)/std 1
<br> solving for x, X=
<br>
<br>C)Same as above. Z score for 5% lowest is -1.65
<br> -1.65 = (X- 30,000)/STD 2, solving for X
<br> X =$392.65
<br>
<br>D)P ( 26,500<= X<= 30,000)
<br> For this Calculate the Z score for both the limits ( 26,500 and 30,000)
<br>
<br>Z1 = (26,500 - 26,500)/STD 1 ;= 0
<br>Z2 = (30,000 - 26,500)/STD 1 ; = .2333
<br>
<br>So % area between Z=0 and Z = .2333 = .5922-.50 = .0922 ( from tables)
<br>Also see the excel sheet.

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