Get the answers you've been looking for with the help of IDNLearn.com's expert community. Our Q&A platform offers reliable and thorough answers to help you make informed decisions quickly and easily.
Sagot :
Answer:
0.8996 = 89.96% probability that the sample proportion will be between 0.2 and 0.4
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]
The manager of a donut store believes that 35% of the customers are first-time customers.
This means that [tex]p = 0.35[/tex]
Sample of 150 customers
This means that [tex]n = 150[/tex]
Mean and standard deviation:
[tex]\mu = p = 0.35[/tex]
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.35*0.65}{150}} = 0.0389[/tex]
What is the probability that the sample proportion will be between 0.2 and 0.4?
p-value of Z when X = 0.4 subtracted by the p-value of Z when X = 0.2.
X = 0.4
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.4 - 0.35}{0.0389}[/tex]
[tex]Z = 1.28[/tex]
[tex]Z = 1.28[/tex] has a p-value of 0.8997
X = 0.2
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.2 - 0.35}{0.0389}[/tex]
[tex]Z = -3.85[/tex]
[tex]Z = -3.85[/tex] has a p-value of 0.0001
0.8997 - 0.0001 = 0.8996
0.8996 = 89.96% probability that the sample proportion will be between 0.2 and 0.4
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com has the answers you need. Thank you for visiting, and we look forward to helping you again soon.
