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lclaudettecarte3550
14.04.2020 •
Mathematics
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 245 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is less than 0.32
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Ответ:
75.17% probability that the sample proportion of households spending more than $125 a week is less than 0.32
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
For the sampling distribution of a proportion p with size n, we have that![\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}](/tpl/images/0598/8594/77fd8.png)
In this problem, we have that:
So
What is the probability that the sample proportion of households spending more than $125 a week is less than 0.32
This is the pvalue of Z when X = 0.32. So
75.17% probability that the sample proportion of households spending more than $125 a week is less than 0.32
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