Question
In: Statistics and Probability
The president of Doerman Distributors, Inc., believes that 26% of the firm's orders come from first-time...
The president of Doerman Distributors, Inc., believes that 26% of the firm's orders come from first-time customers. A simple random sample of 100 orders will be used to estimate the proportion of first-time customers. p=0.26
1. What is the probability that the sample proportion will be between .12 and .40 (to 4 decimals)?
2. What is the probability that the sample proportion will be between .17 and .35 (to 4 decimals)?
Solutions
Expert Solution
Solution
Given that,
p = 0.26
1 - p = 1-0.26 = 0.74
n = 100
= p = 0.26
= 
1) P(0.12 <
< 0.40)
= P[(0.12 -0.26) /0.0439 < (
-
) /
< (0.40 - 0.26) /0.0439 ]
= P( -3.19< z < 3.19)
= P(z <3.19 ) - P(z < -3.19)
= 0.9993 - 0.0007 = 0.9986
probability = 0.9986
2)
P(0.17 <
< 0.35)
= P[(0.17 -0.26) /0.0439 < (
-
) /
< (0.35 - 0.26) /0.0439 ]
= P( -2.05< z < 2.05)
= P(z <2.05 ) - P(z < -2.05)
= 0.9798 - 0.0202= 0.9596
probability = 0.9986
orchestra answered 1 year agoRelated Solutions
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