Mathematics, 26.03.2021 05:50 rhonda45801

g Based on historical data, your manager believes that 32% of the company's orders come from first-time customers. A random sample of 146 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is greater than than 0.43

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Answer from: arangoaiden71

0.0022 = 0.22% probability that the sample proportion is greater than than 0.43

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 zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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 $s = \sqrt{\frac{p(1-p)}{n}}$

Based on historical data, your manager believes that 32% of the company's orders come from first-time customers.

This means that p = 0.32

Mean and standard deviation:

Sample of 146 means that n = 146

$\mu = p = 0.32$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.32*0.68}{146}} = 0.0386$

What is the probability that the sample proportion is greater than than 0.43?

This is 1 subtracted by the pvalue of Z when X = 0.43. So

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.43 - 0.32}{0.0386}$

Z = 2.85

Z = 2.85 has a pvalue of 0.9978

1 - 0.9978 = 0.0022

0.0022 = 0.22% probability that the sample proportion is greater than than 0.43

Answer from: Quest

not only in mathematics, scientific proof also are required in the other field of science. also maths proofs are of particular structure, containing steps and logics. so proving is like doing science, it is science. that's why it is important to prove.

an inaccurate proof leads to errors, or stating results that are not accurate. which wil mess the probelm up you dont want this

Answer from: Quest

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