128,796 views
5 votes
5 votes
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.

User Ganeshja
by
1.9k points

1 Answer

7 votes
7 votes

Answer:

The probability that the mean life expectancy of the sample is less than X years is the p-value of
Z = (X - \mu)/((\sigma)/(√(n))), in which
\mu is the mean life expectancy,
\sigma is the standard deviation and n is the size of the sample.

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 = (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 = (\sigma)/(√(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 = (X - \mu)/((\sigma)/(√(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 = (X - \mu)/((\sigma)/(√(n))), in which
\mu is the mean life expectancy,
\sigma is the standard deviation and n is the size of the sample.

User TachyonicBytes
by
2.2k points