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Mrac [35]
4 years ago
5

Does anyone know the answer to the question

Mathematics
1 answer:
omeli [17]4 years ago
7 0

Answer:

Step-by-step explanation:

Not specifically. That is I don't know what letter the answer is because we can't see all the choices. But I can tell you which one to pick.

Pick the one that looks like the graph below. Do it by where the graph goes up (in quadrant 1) and the y intercept which is 2. Make sure you understand where the 2 came from.

2^x = 2^0 which is 1.

The graph has a y intercept of 2 because the original question is 2*2^x

When x = 0, the intercept = 2*1 = 2

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Answer:

92.65% probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

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 \sigma, the zscore of a measure X is given by:

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

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 pvalue, 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 \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

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

For sums the theorem can also be used, with mean n*\mu and standard deviation s = \sqrt{n}*\sigma

In this problem, we have that:

n = 47, \mu = 47*167 = 7849, s = \sqrt{47}*35 = 240

Use this information to estimate the probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

This is 1 subtracted by the pvalue of Z when X = 7500. So

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

By the Central Limit Theorem

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

Z = \frac{7500 - 7849}{240}

Z = -1.45

Z = -1.45 has a pvalue of 0.0735

1 - 0.0735 = 0.9265

92.65% probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

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3 years ago
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Step-by-step explanation:

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