Question

It is known that 12% of children are nearsighted. A random sample of 170 children is selected. What is the probability more than 14% of this sample will be nearsighted?

	a.	0.7881
	b.	0.2119
	c.	0.2266
	d.	0.7734

Answer 1

P(z<Z) table :

proportion distribution :

mean p = 0.12

SD = [p*(1-p)/n]^0.5

= [0.12*(1-0.12)/170]^0.5

= 0.0249

P(p>0.14) :

{z rounded to 2 decimal = 0.80}

P(x>0.14) = 1 - P(z<0.80)

= 1 - 0.7881

= 0.2119 (option B)

(please UPVOTE)