Based upon statistical studies it has been found that 422 of

Based upon statistical studies it has been found that 4.22% of all households in the United States in 2010 had a combined household income above $250,000. If 14,000 households from 2010 are selected at random, what is the probability that:

a) between 600 and 650 of them (inclusive) had a household income above $250,000?

b) at least 575 of them had a household income above $250,000?

Solution

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    599.5      
x2 = upper bound =    650.5      
u = mean = np =    590.8      
          
s = standard deviation = sqrt(np(1-p)) =    23.7879852      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.365730848      
z2 = upper z score = (x2 - u) / s =    2.509670302      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.642717037      
P(z < z2) =    0.993957804      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.351240766   [ANSWER]

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b)

We first get the z score for the critical value:          
          
x = critical value =    574.5      
u = mean = np =    590.8      
          
s = standard deviation = sqrt(np(1-p)) =    23.7879852      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -0.685219865      
          
Thus, the left tailed area is          
          
P(z >   -0.685219865   ) =    0.753397405 [ANSWER]