Get the information you need from a community of experts on IDNLearn.com. Our community provides accurate and timely answers to help you understand and solve any issue.
Sagot :
Answer:
0.0035 = 0.35% probability that the mean amplifier output would be greater than 432.1 watts.
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.
Mean output of 429 watts with a standard deviation of 10 watts.
This means that [tex]\mu = 429, \sigma = 10[/tex]
Sample of 76:
This means that [tex]n = 76, s = \frac{10}{\sqrt{76}} = 1.147[/tex]
What is the probability that the mean amplifier output would be greater than 432.1 watts?
This is 1 subtracted by the pvalue of Z when X = 432.1. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{432.1 - 429}{1.147}[/tex]
[tex]Z = 2.7[/tex]
[tex]Z = 2.7[/tex] has a pvalue of 0.9965
1 - 0.9965 = 0.0035
0.0035 = 0.35% probability that the mean amplifier output would be greater than 432.1 watts.
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Discover the answers you need at IDNLearn.com. Thank you for visiting, and we hope to see you again for more solutions.
