25 of consumers read the ingredient listed on a products lab

25% of consumers read the ingredient listed on a product\'s label.

For a sample of 220 consumers, what is the probability that between 44 and 68 of them read the ingredients listed on a product\'s label?

Solution

Normal Distribution
Mean ( np ) =0.25 *220 = 55
Standard Deviation ( npq )= 220*0.25*0.75 = 6.4226
Normal Distribution = Z= X- u / sd                   
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 44) = (44-55)/6.4226
= -11/6.4226 = -1.7127
= P ( Z <-1.7127) From Standard Normal Table
= 0.04338
P(X < 68) = (68-55)/6.4226
= 13/6.4226 = 2.0241
= P ( Z <2.0241) From Standard Normal Table
= 0.97852
P(44 < X < 68) = 0.97852-0.04338 = 0.9351