A recent report in USA Today indicated a typical family of f

A recent report in USA Today indicated a typical family of four spends $490 a month on food. Assume the distribution of food expenditures for a family of four follows the normal distributio, with a standard deviation of $90 per month.
a.     What percent of the families spend more than $30 but less than $490 per month on food?
b.     What Percent of the families spend less than $430 per month on food?
c.     What percent spend between $430 and $600 per month on food?
d.     What percent spend between $500 and $600 per month on food?

For a. I have

z=(30-490)/90

z= -5.11

but I don\'t know what to do with this z-value, or how to proceed to solve the rest of the problems. Thank you in advance for any assistance.

Solution

a.

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    30      
x2 = upper bound =    490      
u = mean =    490      
          
s = standard deviation =    90      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -5.111111111      
z2 = upper z score = (x2 - u) / s =    0      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    1.60135E-07      
P(z < z2) =    0.5      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.49999984 = 50.00%      
      

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b.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    430      
u = mean =    490      
          
s = standard deviation =    90      
          
Thus,          
          
z = (x - u) / s =    -0.666666667      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -0.666666667   ) =    0.252492538 = 25.249%

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c.

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    430      
x2 = upper bound =    600      
u = mean =    490      
          
s = standard deviation =    90      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.666666667      
z2 = upper z score = (x2 - u) / s =    1.222222222      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.252492538      
P(z < z2) =    0.889188199      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.636695661 = 63.67%

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d.

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    500      
x2 = upper bound =    600      
u = mean =    490      
          
s = standard deviation =    90      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.111111111      
z2 = upper z score = (x2 - u) / s =    1.222222222      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.544235881      
P(z < z2) =    0.889188199      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.344952318   = 34.495%  

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