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djverab [1.8K]
3 years ago
9

Find area of the figure. Round to the nearest hundredth if necessary.

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
8 0
<span><span>
It would be 17.64 for the area</span></span>
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Read 2 more answers
A population has a mean of 180 and a standard deviation of 24. A sample of 100 observations will be taken. The probability that
jeka94

Answer:

The probability that the mean from that sample will be between 183 and 186 is 0.0994 = 9.94%.

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 \mu and standard deviation \sigma, the z-score 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 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 \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.

A population has a mean of 180 and a standard deviation of 24.

This means that \mu = 180, \sigma = 24

A sample of 100 observations will be taken.

This means that n = 100, s = \frac{24}{\sqrt{100}} = 2.4

The probability that the mean from that sample will be between 183 and 186 is:

This is the pvalue of Z when X = 186 subtracted by the pvalue of Z when X = 183. So

X = 186

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

By the Central Limit Theorem

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

Z = \frac{186 - 180}{2.4}

Z = 2.5

Z = 2.5 has a pvalue of 0.9938

X = 183

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

Z = \frac{183 - 180}{2.4}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

0.9938 - 0.8944 = 0.0994

The probability that the mean from that sample will be between 183 and 186 is 0.0994 = 9.94%.

8 0
3 years ago
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