2 A survey of the entire population of employees conducted b

2. A survey of the entire population of employees conducted by a major car manufacturer found that 53% were concerned about their health benefits. The company now samples 80 employees who took this survey and for some follow-up questions. a. What is the standard deviation of the sample proportion who are concerned about health benefits?

b. What is the probability that the sample proportion is less than 50%

c. What is the upper limit of the sample proportion such that only 3% of the time the sample proportion would exceed this value?

Solution

a)
Proportion ( P ) =0.53
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.53*0.47/80) = 0.0558
b)
P(X < 0.5) = (0.5-0.53)/0.0558
= -0.03/0.0558= -0.5376
= P ( Z <-0.5376) From Standard Normal Table
= 0.2954                  
c)
P ( Z > x ) = 0.03
Value of z to the cumulative probability of 0.03 from normal table is 1.88
P( x-u/ (s.d) > x - 0.53/0.0558) = 0.03
That is, ( x - 0.53/0.0558) = 1.88
--> x = 1.88 * 0.0558+0.53 = 0.635