Question
In: Statistics and Probability
The president of Doerman Distributors, Inc., believes that 30% of the firm’s orders come from first-time...
The president of Doerman Distributors, Inc., believes that 30% of the firm’s orders come from first-time customers. A random sample of 100 orders will be used to estimate the proportion of first-time customers. Assume that the president is correct and p =0.30. What is the probability that the sample proportion will be between 0.25 and 0.35?
Solutions
Expert Solution
Solution
Given that,
p = 0.30
1 - p = 1-0.30=0.70
n = 100
= p =0.30
= 
= P( 0.25 <<0.35
)= P[( 0.25 -0.30) /0.04583 < (
-
) /
< (0.35-0.30) /0.04583 ]
= P( -1.09< z < 109)
= P(z <1.09 ) - P(z <-1.09 )
Using z table
=0.8621-0.1379
=0.7242
probability= 0.7242
orchestra answered 2 years agoRelated Solutions
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