According to a poll on consumer behavior 36 of Americans say

According to a poll on consumer behavior, 36% of American\'s say they will only consider new cars made by an American company when buying one. If you set a random sampling of 200 American\'s:

a. What is the probability that the sample will have between 30-40% who say they will only consider buying a new car from an American company?

b. The probability is 90% that the sample percentage will be contained within what symmetrical limits of the population percentage?

c. The probability is 95% that the sample percentage will be contained within what symmetrical limits of the population percentage?

Solution

Normal Distribution
Proportion ( P ) =0.36
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.36*0.64/200)= 0.0339
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.3) = (0.3-0.36)/0.0339
= -0.06/0.0339 = -1.7699
= P ( Z <-1.7699) From Standard Normal Table
= 0.03837
P(X < 0.4) = (0.4-0.36)/0.0339
= 0.04/0.0339 = 1.1799
= P ( Z <1.1799) From Standard Normal Table
= 0.88099
P(0.3 < X < 0.4) = 0.88099-0.03837 = 0.8426                  
b)
P ( Z < x ) = 0.9
Value of z to the cumulative probability of 0.9 from normal table is 1.282
P( x-u/s.d < x - 0.36/0.0339 ) = 0.9
That is, ( x - 0.36/0.0339 ) = 1.28
--> x = 1.28 * 0.0339 + 0.36 = 0.4035                  
c)
P ( Z < x ) = 0.95
Value of z to the cumulative probability of 0.95 from normal table is 1.645
P( x-u/s.d < x - 0.36/0.0339 ) = 0.95
That is, ( x - 0.36/0.0339 ) = 1.64
--> x = 1.64 * 0.0339 + 0.36 = 0.4158