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?
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
Get Answers For Free
Most questions answered within 1 hours.