Suppose that past records indicate that the probability that

Suppose that past records indicate that the probability that a new car will need a warranty repair in the first 90 days of use is 0.04. If a random sample of 400 new cars is selected, what is the probability that the proportion of new cars needing a warranty repair in the first 90 days will be

(a) between 0.04 and 0.06 ?

(b) above 0.05 ?

(c) If a sample size of 300 is selected, what would be your answers in (a) and (b) ?

Solution

Normal Distribution
Proportion ( P ) =0.04
Standard Deviation = Sqrt (P*Q /n) = 0.04*0.96/400
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.04) = (0.04-0.04)/0.0098
= 0/0.0098 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 0.06) = (0.06-0.04)/0.0098
= 0.02/0.0098 = 2.0408
= P ( Z <2.0408) From Standard Normal Table
= 0.97937
P(0.04 < X < 0.06) = 0.97937-0.5 = 0.4794                  
b)
P(X > 0.05) = (0.05-0.04)/0.0098
= 0.01/0.0098 = 1.0204
= P ( Z >1.02) From Standard Normal Table
= 0.1538                  
c)              
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.04) = (0.04-0.04)/0.0113
= 0/0.0113 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 0.06) = (0.06-0.04)/0.0113
= 0.02/0.0113 = 1.7699
= P ( Z <1.7699) From Standard Normal Table
= 0.96163
P(0.04 < X < 0.06) = 0.96163-0.5 = 0.4616                  

P(X > 0.05) = (0.05-0.04)/0.0113
= 0.01/0.0113 = 0.885
= P ( Z >0.885) From Standard Normal Table
= 0.1881