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
=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
You've reached the end of your free preview.
Want to read all 6 pages?
- Winter '15