It is estimated that 53% of adults in the Lebanese population have never smoked.?

Use the formula

P(X) = 8a +8b +99999c +08637421d

then go to your school, and ask your teacher

a) Proportion of non-smokers = 0.53

Mean of sample proportion = 0.53

Standard deviation of sample proportion = sqrt [ p(1-p)/n] = sqrt [ (0.53)(0.47) / 110] = 0.0476

P( p^ < 0.50) is required :

μ = 0.53

σ = 0.0476

standardize x to z = (x - μ) / σ

P(x < 0.5) = P( z < (0.5-0.53) / 0.0476)

= P(z < -0.6303) = 0.2643

(From Normal probability table)

b)

The z-value for 99% confidence interval is 2.575

Sample proportion phat = 0.6

Variance of proportion = p*(1-p)/n

= 0.6(0.4)/110 =0.0021818

S.D. of p is sqrt[0.002182] = 0.0467

Confidence interval:

phat-zval*sd = 0.6 - (2.575)(0.04671)

phat-zval*sd = 0.6 + (2.575)(0.04671)

99% confidence interval is ( 0.48 , 0.72 )