Question

According to a survey in a​ country, 38% of adults do not own a credit card. Suppose a simple random sample of 600 adults is obtained. Complete parts​ (a) through​ (d) below.

(b) What is the probability that in a random sample of 600adults, more than 42% do not own a credit​ card?

​(c) What is the probability that in a random sample of 600 adults,between 37% and 42% do not own a credit​ card?

Answer 1

Solution:

Given that,

n = 600

a ) = 38% =0.07

1 - = 1 - 0.38 = 093

=   = 0.07

= ( 1 - ) / n

=   0.38 * 0.62 / 600

= 0.0198

= 0.0198

b ) P(   > 0.42 )

= 1 - P (   < 0.42 )

= 1 - P ( -    / ) < ( 0.42 - 0.38 / 0.0198 )

= 1 - P ( z < 0.04 / 0.0198 )

= 1 - P ( z < 2.02 )

Using z table

= 1 - 0.9783

= 0.0217

Probability = 0.0217

c ) P( 0.37 < < 0.42 )  

P( 0.37 - 0.38 / 0.0198 ) < ( -    / ) < ( 0.42 - 0.38 / 0.0198 )

P ( - 0.01 /0.0198 < z < 0.04 / 0.0198 )

P ( - 0.50 < z < 2.02 )

P ( z < 2.02 ) - p ( z < -0.50 )

Using z table

= 0.9783 - 0.3085

= 0.6698

Probability = 0.6698