Question

Based on historical data, your manager believes that 37% of the company's orders come from first-time customers. A random sample of 225 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.26 and 0.38

Answer 1

Answer:

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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 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 $s = \sqrt{\frac{p(1-p)}{n}}$

37% of the company's orders come from first-time customers.

This means that $p = 0.37$

A random sample of 225 orders will be used to estimate the proportion of first-time-customers.

This means that $n = 225$

Mean and standard deviation:

$\mu = p = 0.37$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.37*0.63}{225}} = 0.0322$

What is the probability that the sample proportion is between 0.26 and 0.38?

This is the pvalue of Z when X = 0.38 subtracted by the pvalue of Z when X = 0.26.

X = 0.38

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

By the Central Limit Theorem

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

$Z = \frac{0.38 - 0.37}{0.0322}$

$Z = 0.31$

$Z = 0.31$ has a pvalue of 0.6217

X = 0.26

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

$Z = \frac{0.26 - 0.37}{0.0322}$

$Z = -3.42$

$Z = -3.42$ has a pvalue of 0.0003

0.6217 - 0.0003 = 0.6214

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38