Bookmarks People Window Help wiley plus.com/edugen/student/mainfruni S Black, Bu

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Bookmarks People Window Help wiley plus.com/edugen/student/mainfruni S Black, Business Statistics, 8e SINESS STATISTICS (ECON 245) BU 19 Gresearch, the average family spends about s237 on electronics (computers, cell phones, to an NRF survey conducted back-to-college spending per student, suppose back-to-college family spending on electronics is normally distributed with a etc) in standard deviation of $54. If a family of a returning college student is randomly selected, what is the probability that (a) They spend less than $160 on back-to-college electronics b) They spend more than $370 on back-to-college electronics? (c) They spend between $120 and $175 on back-to-college electronics (Round the values of z to 2 decimal places, Round your answers to 4 decimal places.) (a) P(x 160) (b) P(x 370) (c) PK 120 x 17s) Question Attempts: o of 1 used SAVE OR LATER SUPMIT ANSANER Copyright C 2000-2016 by John Wien & Sons, Inc or related companies. Ai rights reserved. i a200s20us lohn WieLA saca un Al Rights Reserved. A Division of 20hn sana. Version 4.17 3.3 MacBook Air

Explanation / Answer

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    160      
u = mean =    237      
          
s = standard deviation =    54      
          
Thus,          
          
z = (x - u) / s =    -1.43      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.43   ) =    0.0764 [ANSWER]

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B)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    370      
u = mean =    237      
          
s = standard deviation =    54      
          
Thus,          
          
z = (x - u) / s =    2.46      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.46   ) =    0.0069 [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    120      
x2 = upper bound =    175      
u = mean =    237      
          
s = standard deviation =    54      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.17      
z2 = upper z score = (x2 - u) / s =    -1.15      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.015      
P(z < z2) =    0.1251      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.1101   [ANSWER]  

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