A group of 1000 students wrote an entrance exam for the Univ

A group of 1000 students wrote an entrance exam for the University of Statistics. The mean score was 62 with a standard deviation of 12. Assuming a Normal Distribution, answer the following questions: 1. What is the probability of a student scoring above 75? 2. What is the probability of a student failing? (i.e. below 50) How many students failed? 3. What is the minimum mark you would need to score to be in the top 10%? 4. What is the minimum mark you would need to score to be in the top 1%? 5. How many people scored below 30? 6. What is the probability of scoring between 60 and 80?

Solution

1.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    75      
u = mean =    62      
          
s = standard deviation =    12      
          
Thus,          
          
z = (x - u) / s =    1.083333333      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.083333333   ) =    0.139330247 [ANSWER]
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2.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    50      
u = mean =    62      
          
s = standard deviation =    12      
          
Thus,          
          
z = (x - u) / s =    -1      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1   ) =    0.158655254 [ANSWER]

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3.

0.158655254*1000 = 15.86 or 16 students failed. [ANSWER]

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4.

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,          
          
where          
          
u = mean =    62      
z = the critical z score =    2.326347874      
s = standard deviation =    12      
          
Then          
          
x = critical value =    89.91617449   [ANSWER]

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5.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    30      
u = mean =    62      
          
s = standard deviation =    12      
          
Thus,          
          
z = (x - u) / s =    -2.666666667      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.666666667   ) =    0.003830381

Thus, 0.003830381*1000 = 3.8 or 4 people are below 30. [ANSWER, 4]

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6.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    60      
x2 = upper bound =    80      
u = mean =    62      
          
s = standard deviation =    12      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.166666667      
z2 = upper z score = (x2 - u) / s =    1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.433816167      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.499376631   [ANSWER]