Question

Given a population in which the probability of success is

pequals=0.65,

if a sample of

300

items is​ taken, then complete parts a and b.

a.	Calculate the probability the proportion of successes in the sample will be between 0.62 and 0.67
b.	Calculate the probability the proportion of successes in the sample will be between 0.62 and 0.67 if the sample size is 100

a. The probability the proportion of successes in the sample will be between

0.62

and

0.67

is

nothing.

Answer 1

Using central limit theorem,

P( < p ) = P( Z < - p / sqrt( p( 1 - p) / n)

a)

P( 0.62 < < 0.67) = P( < 0.67) - P( < 0.62)

= P( Z < 0.67 - 0.65 / sqrt( 0.65 * 0.35 / 300) - P( Z < 0.62 - 0.65 / sqrt( 0.65 * 0.35 / 300)

= P( Z < 0.7263) - P( Z < -1.0894)

= 0.7662 - 0.1380

= 0.6282

b)

For n = 100

P( 0.62 < < 0.67) = P( < 0.67) - P( < 0.62)

= P( Z < 0.67 - 0.65 / sqrt( 0.65 * 0.35 / 100) - P( Z < 0.62 - 0.65 / sqrt( 0.65 * 0.35 / 100)

= P( Z < 0.4193) - P (Z < -0.6290)

= 0.6625 - 0.2647

= 0.3978