SOLUTION: A Food Marketing Institute found that 30% of households spend more than $125 a week on groceries. Assume the population proportion is 0.3 and a simple random sample of 86 household
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Question 1169956: A Food Marketing Institute found that 30% of households spend more than $125 a week on groceries. Assume the population proportion is 0.3 and a simple random sample of 86 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.37 and 0.49?
Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! z=(0.37-0.30)/sqrt(0.3*0.7/86)
=0.07/0.0494=1.4166
z=(0.49-0.30)/0.0494=3.8462
probability z is between those two values is 0.0782
note, rounding to four decimal places in the above assumes the proportion given will be an integer. It is not, so that the proportion used is a rounding, and four decimal places is not going to be accurate even though it may be calculated.
