A manufacturer of compactdisk players subjects the equipment

A manufacturer of compact-disk players subjects the equipment to a comprehensive testing process for all mechanical and electrical functions before the equipment leaves the factory. Ideally, the hope is that each compact-disk player passes on the first test. Suppose that past data indicate that the probability that a compact-disk player passes the first test is .90.

If 200 players are randomly selected, what is the approximate probability that

a. at least 190 pass the first test?

b. between 175 and 190 pass the first test?

c. fewer than 185 pass the first test?

Solution

Normal Approximation to Binomial Distribution
Mean ( np ) =200 * 0.9 = 180
Standard Deviation ( npq )= 200*0.9*0.1 = 4.2426
Normal Distribution = Z= X- u / sd   
a)
P(X < 190) = (190-180)/4.2426
= 10/4.2426= 2.357
= P ( Z <2.357) From Standard NOrmal Table
= 0.9908                  

P( atleast 190) = 1 - P(X<190) = 1 - 0.9908 = 0.0092
b)
P(X < 185) = (185-180)/4.2426
= 5/4.2426= 1.1785
= P ( Z <1.1785) From Standard NOrmal Table
= 0.8807                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 175) = (175-180)/4.2426
= -5/4.2426 = -1.1785
= P ( Z <-1.1785) From Standard Normal Table
= 0.11929
c)
P(X < 190) = (190-180)/4.2426
= 10/4.2426 = 2.357
= P ( Z <2.357) From Standard Normal Table
= 0.99079
P(175 < X < 190) = 0.99079-0.11929 = 0.8715