Question

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

Answer 1

Answer:

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