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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 and the standard deviation was years. If a sample of people from this region is selected, find the probability that the mean life expectancy will be less than years. Round intermediate -value calculations to decimal places and round the final answer to at least decimal places.

Sagot :

Answer:

The probability that the mean life expectancy of the sample is less than X years is the p-value of [tex]Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex], in which [tex]\mu[/tex] is the mean life expectancy, [tex]\sigma[/tex] 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 [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 establishes 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.

We have:

Mean [tex]\mu[/tex], standard deviation [tex]\sigma[/tex].

Sample of size n:

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

[tex]Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

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 [tex]Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex], in which [tex]\mu[/tex] is the mean life expectancy, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.

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