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bezimeni [28]
3 years ago
10

A Food Marketing Institute found that 34% of households spend more than $125 a week on groceries. Assume the population proporti

on 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
Mathematics
1 answer:
sammy [17]3 years ago
7 0

Answer:

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 \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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 \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

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 \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

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

Mean and standard deviation:

\mu = p = 0.34

s = \sqrt{\frac{0.34*0.66}{124}} = 0.0425

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:

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.31 - 0.34}{0.0425}

Z = -0.705

Z = -0.705 has a p-value of 0.2405.

0.2405 = 24.05% probability that the sample proportion of households spending more than $125 a week is less than 0.31.

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