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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)) = P(Z < [x-mean]/S)

P(X>x) = P(Z > x -mean/std/sqrt(n)) = P(Z >[x-mean]/S)

P(y<X<x) = P(X<x) - P(X<y)

mean = 67%

std = standard deviation = 16%

sample standard deviation S = 16%/sqrt(16) = 4%

n = size of sample = 16

a)

P(X>70%) = P(Z > (70% - 67%)/4%)

= P(Z>0.75)

= 0.2266 (from normal distribution table)

b)

P(X < 60%) = P(Z < (60% -67%)/4%)

= 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