Question

Suppose a simple random sample of size n=200 is obtained from a population whose size is Upper N= 20,000 and whose population proportion with a specified characteristic is

p equals 0.6 .p=0.6. Complete parts ​(a) through​ (c) below.

(a) Determine the standard deviation

(b) What is the probability of obtaining x=124 or more individuals with the​ characteristic? That​ is, what is

​P(p≥0.62)?

(c) What is the probability of obtaining x=106 or fewer individuals with the​ characteristic? That​ is, what is

​P(p ≤0.53)?

Answer 1

Solution

Given that,

p = 0.60

1 - p = 1 - 0.60 = 0.40

n = 200

a) = p = 0.60

=  [p( 1 - p ) / n] = [(0.60* 0.40) / 200 ] = 0.0346

b) P( ≥ 0.62 ) = 1 - P( ≤ 0.62 )

= 1 - P(( - ) / ≤ ( 0.62 - 0.60) / 0.0346)

= 1 - P(z ≤ 0.58 )

Using z table

= 1 - 0.7190

= 0.2810

c) ​P( ≤ 0.53 )

= P[( - ) / ≤ ( 0.53 - 0.60 ) / 0.0346 ]

= P(z ≤ -2.02 )

Using z table

= 0.0217