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Tanzania [10]
3 years ago
15

How Many Periods of time,each 1/3 of an hour long,does a 8 hour period represent

Mathematics
1 answer:
bonufazy [111]3 years ago
6 0
24 I'm pretty sure because 3 periods per hour for 8 hours so 8x3=24
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Which shows the correct way to convert 5 ounces to pounds
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There are 16 ounces in 1 lb. 

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The mean output of a certain type of amplifier is 183 watts with a variance of 121. If 78 amplifiers are sampled, what is the pr
koban [17]

Answer:

0.8278 = 82.78% probability that the mean of the sample would differ from the population mean by less than 1.7 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 normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation(which is the square root of the variance) \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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Central Limit Theorem

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 183, \sigma = \sqrt{121} = 11, n = 78, s = \frac{11}{\sqrt{78}} = 1.2455

Probability that the mean of the sample would differ from the population mean by less than 1.7 watts?

This is the pvalue of Z when X = 183+1.7 =184.7 subtracted by the pvalue of Z when X = 183 - 1.7 = 181.3.

X = 184.7

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{184.7 - 183}{1.2455}

Z = 1.365

Z = 1.365 has a pvalue of 0.9139

X = 181.3

Z = \frac{X - \mu}{s}

Z = \frac{181.3 - 183}{1.2455}

Z = -1.365

Z = -1.365 has a pvalue of 0.0861

0.9139 - 0.0861 = 0.8278

0.8278 = 82.78% probability that the mean of the sample would differ from the population mean by less than 1.7 watts

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