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The president of Doerman Distributors, Inc., believes that 30% of the firm's orders come from first-time customers. A random sample of 150 orders will be used to estimate the proportion of first-time customers.Required:a. Assume that the president is correct and p = 0.30. What is the sampling distribution of p bar for this study?b. What is the probability that the sample proportion p bar will be between 0.20 and 0.40?c. What is the probability that the sample proportion will be between 0.25 and 0.35?

Respuesta :

Answer:

Step-by-step explanation:

p = .3

n = 150

p(bar ) = 1 - p = .7

[tex]\sigma_p=\sqrt\frac{p(1-p) }{n} }[/tex]

[tex]\sigma_p=\sqrt\frac{.3(1-.3) }{150} }[/tex]

=.037

b )

P ( .2 <p<.4 ) = P [ (.2 - .3) / .037 < z < ( .4 - .3 ) / .037 ]

= P [ (-2.7  < z < +2.7 ]

= .9965-.0035

= .993

c )

P ( .25 <p<.35 ) = P [ (.25 - .3) / .037 < z < ( .35 - .3 ) / .037 ]

= P [ (-1.35  < z < +1.35 ]

= .9115 - .0885

= .823

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