Although controversial and the subject of some recent law suits, (e.g., Satchell

Question

Although controversial and the subject of some recent law suits, (e.g., Satchell et al. vs. FedEx), some human resource departments administer standard IQ tests to all employees. The StanfordBinet test scores are well modeled by a normal distribution with a mean of 100 and a standard deviation of 16.

Question. If the test is given to 121 job applicants, what is the probability that between 22 and 33 applicants will have a test score greater than 111?

the answer is 0.7085 and I want to know how to get this answer

Explanation / Answer

Mean ( u ) =100
Standard Deviation ( sd )=16
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X > 111) = (111-100)/16
= 11/16 = 0.6875
= P ( Z >0.688) From Standard Normal Table
= 0.2451                  

and now,
P = 0.2451,
n = 121

mean ( np ) =121 * 0.2451 = 29.6571
standard deviation ( npq )= 121*0.2451*0.7549 = 4.7316
Normal Distribution = Z= X- u / sd                   
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 22) = (22-29.657)/4.7316
= -7.657/4.7316 = -1.6183
= P ( Z <-1.6183) From Standard Normal Table
= 0.0528
P(X < 33) = (33-29.657)/4.7316
= 3.343/4.7316 = 0.7065
= P ( Z <0.7065) From Standard Normal Table
= 0.76007
P(22 < X < 33) = 0.76007-0.0528 = 0.7085  

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