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15.07.2021 •
Mathematics
A Food Marketing Institute found that 34% of households spend more than $125 a week on groceries. Assume the population proportion is 0.34 and a simple random sample of 124 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 0.31
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Ответ:
0.2405 = 24.05% probability that the sample proportion of households spending more than $125 a week is less than 0.31.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
and standard deviation ![s = \sqrt{\frac{p(1-p)}{n}}](/tpl/images/1394/6240/73804.png)
Assume the population proportion is 0.34 and a simple random sample of 124 households is selected from the population.
This means that![p = 0.34, n = 124](/tpl/images/1394/6240/554ff.png)
Mean and standard deviation:
What is the probability that the sample proportion of households spending more than $125 a week is less than 0.31?
This is the p-value of Z when X = 0.31, so:
0.2405 = 24.05% probability that the sample proportion of households spending more than $125 a week is less than 0.31.
Ответ:
Can i please have a Brainlier