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

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Answer:

This probability that the sample proportion is between 0.35 and 0.45 is the p-value of [tex]Z = \frac{0.45 - \mu}{\sqrt{\frac{p(1-p)}{160}}}[/tex] subtracted by the p-value of [tex]Z = \frac{0.35 - \mu}{\sqrt{\frac{p(1-p)}{160}}}[/tex]. 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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

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

Estimate of the proportion:

This is p, and thus, we use [tex]\mu = p[/tex].

Sample of 160

This means that:

[tex]n = 160, s = \sqrt{\frac{p(1-p)}{160}}[/tex]

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 [tex]Z = \frac{0.45 - \mu}{\sqrt{\frac{p(1-p)}{160}}}[/tex] subtracted by the p-value of [tex]Z = \frac{0.35 - \mu}{\sqrt{\frac{p(1-p)}{160}}}[/tex]. These p-values are found looking at the z-table.