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# A Food Marketing Institute found that 51% of households spend more than S125 a week on groceries: Assume the population proportion is 0.51 and a simple random sampl...

## Question

###### A Food Marketing Institute found that 51% of households spend more than S125 a week on groceries: Assume the population proportion is 0.51 and a simple random sample of 60 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is more than than 0.46?Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations_Answer(Enter your answer as a number A Food Marketing Institute found that 51% of households spend more than S125 a week on groceries: Assume the population proportion is 0.51 and a simple random sample of 60 households is selected from the population. What is the probability that the sample proportion of households spending more than$125 a week is more than than 0.46? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations_ Answer (Enter your answer as a number accurate to 4 decimal places. leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 15 weeks. Assume that for the population of all unemployed individuals the length of unemployment is normally distributed where the population mean length of unemployment is 15 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 39 unemployed individuals for a follow-up study. Find the probability that a single randomly selected value is less than 14. P(X < 14) Find the probability that a sample of size n P(M 14) 39 is randomly selected with a mean less than 14. Enter your answers as numbers accurate to 4 decimal places.  #### Similar Solved Questions

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