According to a 2009 Reader's Digest article, people throw away about 12% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 161 grocery shoppers to investigate their behavior. What is the probability that the sample proportion does not exceed 0.09?
Solution
Given that,
p = 0.12
1 - p = 1 -0.12 =0.88
n = 161
= p =0.12
= [p ( 1 - p ) / n] = [(0.12*0.88) /161 ] = 0.0256
P( <0.09 ) =
= P[( - ) / < (0.09 -0.12) /0.0256 ]
= P(z <-1.17 )
Using z table,
=0.1210
probability=0.1210
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