please help will rate thank you The average amount parents s

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The average amount parents spent per child on back-to-school items in August 2012 was $744 with a standard deviation of $265. Assume the amount spent on back-to-school items is normally distributed.

A) Find the probability the amount spent on a randomly selected child is greater than $313.

B) Find the probability the amount spent on a randomly selected child is less than $600 or greater than or equal to $950.

C) Find the probability the amount spent on a randomly selected child is less than or equal to $768.

D) Find the probability the amount spent on a randomly selected child is greater than $177 and less than $787.

E) The probability is 0.42 that the amount spent on a randomly selected child will be between what two values equidistant from the mean? Find the lower endpoint and upper endpoint.

F) The probability is 0.77 that the amount spent on a randomly selected child is at most what value?

G) The probability is 0.86 that the amount spent on a randomly selected child is no less than what value?

Solution

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    313      
u = mean =    744      
          
s = standard deviation =    265      
          
Thus,          
          
z = (x - u) / s =    -1.626415094      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1.626415094   ) =    0.948069309 [ANSWER]

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B)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    600      
x2 = upper bound =    950      
u = mean =    744      
          
s = standard deviation =    265      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.543396226      
z2 = upper z score = (x2 - u) / s =    0.777358491      
          
Using table/technology, the left tailed areas between these z scores is          
          
          
P(z < z2) =    0.781526352      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.781526352      

Thus, those outside this interval is the complement = 0.218473648   [ANSWER]

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C)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    768      
u = mean =    744      
          
s = standard deviation =    265      
          
Thus,          
          
z = (x - u) / s =    0.090566038      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   0.090566038   ) =    0.536081291 [ANSWER]

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D)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    177      
x2 = upper bound =    787      
u = mean =    744      
          
s = standard deviation =    265      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.139622642      
z2 = upper z score = (x2 - u) / s =    0.162264151      
          
Using table/technology, the left tailed areas between these z scores is          
          
          
P(z < z2) =    0.564451078      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.564451078   [answer]  

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