Question

1. Suppose that 31% of adults have at least one tattoo. If you sample 90 random adults, what is the probability that 33% or more of them have a tattoo?

2. Instead, if you sample 300 random adults, what is the probability that 33% or more of them have a tattoo?

Answer 1

Solution

Given that,

p = 0.31

1 - p = 1 - 0.31 = 0.69

1) n = 90

= p = 0.31

=  [p( 1 - p ) / n] = [(0.31 * 0.69) / 90 ] = 0.0488

P(    0.33) = 1 - P( 0.33 )

= 1 - P(( - ) /    (0.33 - 0.31) / 0.0488)

= 1 - P(z 0.41)

= 1 - 0.6591

= 0.3409

2) n = 300

= p = 0.31

=  [p( 1 - p ) / n] = [(0.31 * 0.69) / 300 ] = 0.0267

P(    0.33) = 1 - P( 0.33 )

= 1 - P(( - ) /    (0.33 - 0.31) / 0.0267)

= 1 - P(z 0.75)

= 1 - 0.7734

= 0.2266