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Xelga [282]
3 years ago
14

What is 28 divide by 3

Mathematics
2 answers:
Juli2301 [7.4K]3 years ago
8 0
9.3333333333333333333333..............

or

-
9.33
yarga [219]3 years ago
5 0

The answer is 9.3

Step-by-step explanation:

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Life Expectancies In a study of the life expectancy of people in a certain geographic region, the mean age at death was years an
Sphinxa [80]

Answer:

The probability that the mean life expectancy of the sample is less than X years is the p-value of Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}, in which \mu is the mean life expectancy, \sigma is the standard deviation and n is the size of the sample.

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 establishes 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.

We have:

Mean \mu, standard deviation \sigma.

Sample of size n:

This means that the z-score is now, by the Central Limit Theorem:

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

Find the probability that the mean life expectancy will be less than years.

The probability that the mean life expectancy of the sample is less than X years is the p-value of Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}, in which \mu is the mean life expectancy, \sigma is the standard deviation and n is the size of the sample.

8 0
2 years ago
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