Given u 35 o22 n120 Ex uSDx aSqr n 22Sqr120 02008 Px3 Px ExSDx 3 3502008 Pz 249

Given u 35 o22 n120 ex usdx asqr n 22sqr120 02008 px3

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Given, u= 3.5 o=2.2 n=120 E(x) = u;SD(x) = a/Sqr n = 2.2/Sqr120 =0.2008 P(x<3)  = P(x-E(x)/SD(x) < 3-3.5/0.2008) =P(z) <-2.49 =0.0064 3. Traditionally, 25% of cell phone company customers leave after their term contract expires.  Assume this is the true population proportion of churn (customers leaving after their term  expired) and answer the following questions: ( 6 points ) a. If simple random samples of 200 customers are taken from the population of  customers of this company, describe the sampling distribution of the sample proportion of  churn. Given P=0.3;n=100 sample proportion (p) – x/n E(p) =1/n E(x)  =Np/n =0.3 Var 9P0 = VAR X/N
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=NPQ/N2 =PQ/N Sd(P) =SQR PQ/N = 0.045826 b. What is the probability that a simple random sample of 200 customers will have a  sample proportion of churn that is within ± 0.04 of the population proportion? P =0.03 POPULATION PORTION = 0.03 0.3 P 0.27 <P <0.33  =P(0.27 0.3/0.045826 < Z 0.33 0.03/0.045826 =p -0.065465 < Z 0.654654  =0.7437 – 0.2563 =0.4873 c. Check the condition for using the normal distribution to approximate the sampling  distribution of the sample proportion and see whether it is valid. THE CONDITION FOR USING THE NORMAL DISTRIBUTION IS ATLEAST 10 SUCCESS  AND 10 FAILURES NUMBER OF SUCCESS = 100 *0.3 = 30 NUMBER OF FAILURES = 100*0.7 = 70 THE CONDITIONS ARE SATISFIED  4. A simple random sample of 100 items is taken from an inventory and the average unit cost  on the items is $25.5. The population standard deviation   is known to be $10.25. ( σ 8 points ) Given u=$25 o=$10.5 n = 120  95% confidence interval for the mean unit cost of the population is Mean = $25 + - 1.96 * 10.5/ sqr 120
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=23.121, 26.879 
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