Question

It is generally believed that nearsightedness affects about 12% of all children. A school district has registered 170 incoming kindergarten children. If a random sample of 50 kindergarten children is selected, what is the probability the sample differs from the mean by more than 1%? Round your answers to four decimal places.

( please show step by step) thanks.

Answer 1

Solution

Given that,

=  [p ( 1 - p ) / n] = [(0.12 * 0.88) / 50 ] = 0.0460

= 1 - P[(-0.01) /0.0460 < ( - ) / < (0.01) / 0.0460]

= 1 - P(-0.22 < z < 0.22)

= 1 - P(z < 0.22) - P(z < -0.22)

= 1 - P(0.5871 - 0.4129)

= 1 - 0.1742

= 0.8258

Probability = 0.8258