A personal shopper has built a business shopping for busy st

A personal shopper has built a business shopping for busy students and charges them a percentage of their purchases. Currently the shopper has 50 clients who spend an approximate normal mean if $200 with a standard deviation of $75 per week.

1) probability of 1 client spending more than $250 in a given week

2) probability 1 client will spend between $125 and $200 in a given week

3) probability that average spending across all 50 clients will exceed $210

4) What is the 10th percentile of average spending across all 50 clients

5) If the shopper charges clients 10% of their purchase amount, what is the probability the shopper will earn more than $1,100 in a week shopping for 50 clients?

Solution

1)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    250      
u = mean =    200      
          
s = standard deviation =    75      
          
Thus,          
          
z = (x - u) / s =    0.666666667      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.666666667   ) =    0.252492538 [answer]

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2)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    125      
x2 = upper bound =    200      
u = mean =    200      
          
s = standard deviation =    75      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1      
z2 = upper z score = (x2 - u) / s =    0      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.5      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.341344746   [ANSWER]

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3)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    210      
u = mean =    200      
n = sample size =    50      
s = standard deviation =    75      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    0.942809042      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.942809042   ) =    0.172889293 [answer]

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4)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.1      
          
Then, using table or technology,          
          
z =    -1.281551566      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    200      
z = the critical z score =    -1.281551566      
s = standard deviation =    75      
n = sample size =    50      
Then          
          
x = critical value =    186.407093   [ANSWER]

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5)

This is just like asking for the probability that the sum of those shopped is $1100/0.1 = $11000.

Thus, like asking that the mean of the 50 shoppers is $11000/50 = $220.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    220      
u = mean =    200      
n = sample size =    50      
s = standard deviation =    75      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.885618083      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.885618083   ) =    0.029673219 [ANSWER]