Alas Ash only got a 73 on his exam but the professor decided

Alas, Ash only got a 73% on his exam, but the professor decided to curve the class based on a normal distribution. Students with a standard deviation above the average will get an A. The average grade on the test is 65% and the standard deviation is 7.3.

a) Did Ash get an A on his exam?

b) If the class size is 100, how many students received an 80% or higher on the exam? (Round down).

c) How many students received a 70% to 80% inclusive on the exam? (Round down).

d) Find the probability of getting exactly a 78% on the exam.

Solution

a)

The cut off for an A is at least

x = u + sigma = 65 + 7.3 = 72.3

As he got 73%, then YES, HE GETS AN A. [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    80      
u = mean =    65      
          
s = standard deviation =    7.3      
          
Thus,          
          
z = (x - u) / s =    2.054794521      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.054794521   ) =    0.019949428

Thus, on a class of 100, there are 0.019949428 = 1.995 students. As we are told to round down, there is just 1 student who got 80 or above.

[If you use tables in class, you will get z = 2.05, and will get an area of 0.0201. Thus, you should answer 2 STUDENTS. [ANSWER]

The answer above is the exact one, without rounding.]

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c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    70      
x2 = upper bound =    80      
u = mean =    65      
          
s = standard deviation =    7.3      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.684931507      
z2 = upper z score = (x2 - u) / s =    2.054794521      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.753306428      
P(z < z2) =    0.980050572      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.226744144      

In a class of 100, there will be 0.226744144*100 = 22.67 or 22, when we round down. [ANSWER]

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d)

Strictly speaking, the probability of a point score is 0. [answer]

But if you assign a width of 1 to a point score in your class, adding/subtracting 0.5 to the point score, we do:

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    77.5      
x2 = upper bound =    78.5      
u = mean =    65      
          
s = standard deviation =    7.3      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    1.712328767      
z2 = upper z score = (x2 - u) / s =    1.849315068      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.956581951      
P(z < z2) =    0.967793834      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.011211883   [answer]