Business Statistics Class The average grade point average GP

Business Statistics Class

The average grade point average (GPA) of undergraduate students in New York is normally distributed with a population mean of 2.5 and a population standard deviation of 0.5. Compute the following, showing all work:

(I) The percentage of students with GPA\'s between 2.0 and 2.3 is:

(II) The percentage of students with GPA\'s below 2.7 is:

(III) Above what GPA will the top 5% of the students be (i.e., compute the 95th percentile):   

(IV) If a sample of 36 students is taken, what is the probability that the sample mean GPA will be between 2.60 and 2.75?  

Solution

i)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    2      
x2 = upper bound =    2.3      
u = mean =    2.5      
          
s = standard deviation =    0.5      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1      
z2 = upper z score = (x2 - u) / s =    -0.4      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.344578258      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.185923004 = 18.5923004% [answer]

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ii)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    2.7      
u = mean =    2.5      
          
s = standard deviation =    0.5      
          
Thus,          
          
z = (x - u) / s =    0.4      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   0.4   ) =    0.655421742 = 65.5421742 [answer]

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iii)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.95      
          
Then, using table or technology,          
          
z =    1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    2.5      
z = the critical z score =    1.644853627      
s = standard deviation =    0.5      
          
Then          
          
x = critical value =    3.322426813   [answer]

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iv)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    2.6      
x2 = upper bound =    2.75      
u = mean =    2.5      
n = sample size =    36      
s = standard deviation =    0.5      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    1.2      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    3      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.88493033      
P(z < z2) =    0.998650102      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.113719772   [answer]