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
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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
and standard deviation ![s = \sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
37% of the company's orders come from first-time customers.
This means that ![p = 0.37](https://tex.z-dn.net/?f=p%20%3D%200.37)
A random sample of 225 orders will be used to estimate the proportion of first-time-customers.
This means that ![n = 225](https://tex.z-dn.net/?f=n%20%3D%20225)
Mean and standard deviation:
![\mu = p = 0.37](https://tex.z-dn.net/?f=%5Cmu%20%3D%20p%20%3D%200.37)
![s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.37*0.63}{225}} = 0.0322](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.37%2A0.63%7D%7B225%7D%7D%20%3D%200.0322)
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}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{0.38 - 0.37}{0.0322}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B0.38%20-%200.37%7D%7B0.0322%7D)
![Z = 0.31](https://tex.z-dn.net/?f=Z%20%3D%200.31)
has a pvalue of 0.6217
X = 0.26
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{0.26 - 0.37}{0.0322}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B0.26%20-%200.37%7D%7B0.0322%7D)
![Z = -3.42](https://tex.z-dn.net/?f=Z%20%3D%20-3.42)
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