Based on past data the sample mean of the credit card purcha

Based on past data, the sample mean of the credit card purchases at a large department store is $35. Assuming sample size is 25 and the population standard deviation is 10.

a)         What % of samples are likely to have between 20 and 30?   _________

b)         Between what two values 90% of sample means fall?                         ____________

c)          Below what value 99% of sample means fall?

d)         Above what value only 1% of sample means fall??

Within what symmetrical limits of the population percentage will 95% of the sample percentages fall? ___________

Solution

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    20      
x2 = upper bound =    30      
u = mean =    35      
n = sample size =    1      
s = standard deviation =    10      
          
Thus, the two z scores are          
          
z1 = lower z score =    -1.5      
z2 = upper z score =    -0.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.066807201      
P(z < z2) =    0.308537539      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.241730337   [ANSWER, A]

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Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
              
X = sample mean =    35          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    10          
n = sample size =    25          
              
Thus,              
              
Lower bound =    31.71029275          
Upper bound =    38.28970725          
              
Thus, the confidence interval is              
              
(   31.71029275   ,   38.28970725   ) [ANSWER, B]

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Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
              
X = sample mean =    35          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    10          
n = sample size =    25          
              
Thus,              
              
Lower bound =    29.84834139          
Upper bound =    40.15165861          
              
Thus, the confidence interval is              
              
(   29.84834139   ,   40.15165861   ) [ANSWER, C]

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First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.99      
          
Then, using table or technology,          
          
z =    2.326347874      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    35      
z = the critical z score =    2.326347874      
s = standard deviation =    10      
n = sample size =    25      
Then          
          
x = critical value =    39.65269575 [ANSWER, D]

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