How do do all parts with a calcuator In an examination the p
How do do all parts with a calcuator
 In an examination the probability distribution of scores (X) can be approximated by normal distribution with mean 62.6 and standard deviation 7.5.
Suppose one has to obtain at least 55 to pass the exam. What is the probability that a randomly selected student passed the exam? 0.8445
If two students are selected randomly what is the chance that both the students failed? 0.0242
If only top 4% students are given an award, then what was the minimum marks required to get the award? 75.7000
| A: 0.0062 | B: 0.25 | C: 0.4651 | D: 0.538 | E: 0.8445 | F: 0.9075 | 
Solution
If only top 4% students are given an award, then what was the minimum marks required to get the award?
First, we get the z score from the given left tailed area. As          
           
 Left tailed area = 1 - 0.04 = 0.96      
           
 Then, using table or technology,          
           
 z =    1.750686071      
           
 As x = u + z * s,          
           
 where          
           
 u = mean =    62.6      
 z = the critical z score =    1.750686071      
 s = standard deviation =    7.5      
           
 Then          
           
 x = critical value =    75.73014553 = 75.7 [ANSWER, D]      

