Question

A candy company claims that its jelly bean mix contains 21​% blue jelly beans. Suppose that the candies are packaged at random in small bags containing about 332 jelly beans. What is the probability that a bag will contain more than 23​% blue jelly​ beans?   

Answer 1

Solution

Given that,

p = 0.21

1 - p = 1 - 0.21=0.79

n = 332

= p =0.21

=  [p( 1 - p ) / n] = [(0.21*0.79) /332 ] = 0.0224

P( >0.23 ) = 1 - P( < 0.23)

= 1 - P(( - ) / < (0.23 - 0.21) /0.0224 )

= 1 - P(z <0.89 )

Using z table

= 1 -0.8133

=0.1867

probability=0.1867