Question

Suppose 56V% of the population has myopia. If a random sample of size 886886 is selected, what is the probability that the proportion of persons with myopia will be less than 59Y%

Answer 1

0.9608

Explanation:

P = 56%

56% = 0.56

1-p

1-0.56

= 0.44

√p(1-p/n

= √0.56(1-0.56)/886

= √0.0002781

= 0.01667

= 0.017

0.59-0.56/0.017

= 1.76

P(z <1.76)

When we check the z table,

Probability is equal to 0.9608