Question

from first-time customers. A random sample of 161 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.35 and 0.45

Answer 1

Answer:

This probability that the sample proportion is between 0.35 and 0.45 is the p-value of $Z = \frac{0.45 - \mu}{\sqrt{\frac{p(1-p)}{160}}}$ subtracted by the p-value of $Z = \frac{0.35 - \mu}{\sqrt{\frac{p(1-p)}{160}}}$. These p-values are found looking at the z-table.

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

Estimate of the proportion:

This is p, and thus, we use $\mu = p$.

Sample of 160

This means that:

$n = 160, s = \sqrt{\frac{p(1-p)}{160}}$

What is the probability that the sample proportion is between 0.35 and 0.45?

This probability that the sample proportion is between 0.35 and 0.45 is the p-value of $Z = \frac{0.45 - \mu}{\sqrt{\frac{p(1-p)}{160}}}$ subtracted by the p-value of $Z = \frac{0.35 - \mu}{\sqrt{\frac{p(1-p)}{160}}}$. These p-values are found looking at the z-table.