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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.
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