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Factor 6 z^3/3 + 1 z^1/3
1 answer:
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Answer:
0.6210
Step-by-step explanation:
Given that a Food Marketing Institute found that 39% of households spend more than $125 a week on groceries
Sample size n =87
Sample proportion will follow a normal distribution with p =0.39
and standard error =
the probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41
=
There is 0.6210 probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41