The College Alcohol Study interviewed an SRS of 14941 colleg

The College Alcohol Study interviewed an SRS of 14,941 college students about their drinking habits. Suppose that 40% of all college students \"drink to get drunk\" at least once in a while. That is, p = 0.4.

Probability for n = 1000:  
Probability for n = 4000:  
Probability for n = 16000:

Solution

Points To Pass for normal approximation:
1) Experiment Consistes of a swequencce of n identical Trials
2) Only 2 Outcomes are possible on each trail, Success or Failure
3) Trials are independent & Below conditions should satisfy                  
b)                  
Proportion ( P ) =0.4
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.4*0.6/14941)
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.39) = (0.39-0.4)/0.004
= -0.01/0.004 = -2.5
= P ( Z <-2.5) From Standard Normal Table
= 0.00621
P(X < 0.41) = (0.41-0.4)/0.004
= 0.01/0.004 = 2.5
= P ( Z <2.5) From Standard Normal Table
= 0.99379
P(0.39 < X < 0.41) = 0.99379-0.00621 = 0.9876                  
  
c) WHEN n=1000
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.4*0.6/1000)=0.0155
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.39) = (0.39-0.4)/0.0155
= -0.01/0.0155 = -0.6452
= P ( Z <-0.6452) From Standard Normal Table
= 0.25941
P(X < 0.41) = (0.41-0.4)/0.0155
= 0.01/0.0155 = 0.6452
= P ( Z <0.6452) From Standard Normal Table
= 0.74059
P(0.39 < X < 0.41) = 0.74059-0.25941 = 0.4812                  

WHEN n=4000
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.4*0.6/4000)=0.0077
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.39) = (0.39-0.4)/0.0077
= -0.01/0.0077 = -1.2987
= P ( Z <-1.2987) From Standard Normal Table
= 0.09702
P(X < 0.41) = (0.41-0.4)/0.0077
= 0.01/0.0077 = 1.2987
= P ( Z <1.2987) From Standard Normal Table
= 0.90298
P(0.39 < X < 0.41) = 0.90298-0.09702 = 0.806                  

WHEN n=16000
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.4*0.6/16000)=0.0039
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.39) = (0.39-0.4)/0.0039
= -0.01/0.0039 = -2.5641
= P ( Z <-2.5641) From Standard Normal Table
= 0.00517
P(X < 0.41) = (0.41-0.4)/0.0039
= 0.01/0.0039 = 2.5641
= P ( Z <2.5641) From Standard Normal Table
= 0.99483
P(0.39 < X < 0.41) = 0.99483-0.00517 = 0.9897