Respuesta :
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 [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 estabilishes 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]
37% of the company's orders come from first-time customers.
This means that [tex]p = 0.37[/tex]
A random sample of 225 orders will be used to estimate the proportion of first-time-customers.
This means that [tex]n = 225[/tex]
Mean and standard deviation:
[tex]\mu = p = 0.37[/tex]
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.37*0.63}{225}} = 0.0322[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.38 - 0.37}{0.0322}[/tex]
[tex]Z = 0.31[/tex]
[tex]Z = 0.31[/tex] has a pvalue of 0.6217
X = 0.26
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.26 - 0.37}{0.0322}[/tex]
[tex]Z = -3.42[/tex]
[tex]Z = -3.42[/tex] 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
