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

=  [p ( 1 - p ) / n] = [0.26*(0.74) /100 ] = 0.0439

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


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