5 The New York Times has reported that 67 of adult US citize

5. The New York Times has reported that 67% of adult US citizens support the government\'s efforts against ISIS with a standard deviation of 16%. A randomly selected group of 16 students are asked about their support of the government\'s efforts. What is the probability that their level of support is a) Greater than 70%? b) Less than 60%? c) Between 60% and 70%?

Solution

Let X be the random variable represents the number of us citizens supporting governments efforts

X follows normal distribution

P(X < x) = P(Z < x -mean/std/sqrt(n))

a)

P(X>70) = P(Z > 70 - 67/16/sqrt(16)) = P(Z>0.75) = 0.2266

b)

P(X < 60) = P(Z < 60 -67/16/sqrt(16)) = P( Z<-1.75) = 0.0401

c)

P(60 < X < 70) = P(X<70) - P(X <60)

= P(Z<0.75) - P(Z < -1.75)

= 0.7734 - 0.0401

= 0.7333