Question

A study investigating the relationship between age and annual medical expenses randomly samples 400 individuals in a city. It is hoped that the sample will have a similar mean age as the entire population. Complete parts a and b below.
a. If the standard deviation of the ages of all individuals in the city is σ=16​,find the probability that the mean age of the individuals sampled is within one year of the mean age for all individuals in the city.​ (Hint: Find the sampling distribution of the sample mean age and use the central limit theorem. You​ don't have to know the population mean to answer​this, but if it makes it​ easier, use a value such as μ=30​.)

The probability is ____

​(Round to three decimal places as​ needed.)

Answer 1

Answer:

Step-by-step explanation:

Hello!

To investigate the relationship between age and annual medical expenses a random sample of 400 individuals was taken.

The sample is expected to be representative of the entire population.

a.

The variable of interest is X: the age of an individual in the city.

The standard deviation is known to be δ= 16

Applying the Central Limit Theorem we can approximate the distribution of the sampling distribution to normal.

Using this approximation you can use the standard normal distribution $Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }$≈N(0;1)

P(μ-1≤X[bar]≤μ+1)= P(X[bar]≤μ+1)-P(X[bar]≤μ-1)

Now using the standard normal distribution you have to standardize both terms:

P(X[bar]≤μ+1)= P(Z≤[(μ+1)-μ]/(σ/√n))= P(Z≤(μ-μ+1)/(15/√400))= P(Z≤1/(15/√400))= P(Z≤1.33)

P(X[bar]≤μ-1)= P(Z≤[(μ-1)-μ]/(σ/√n))= P(Z≤(μ-μ-1)/(15/√400))= P(Z≤-1/(15/√400))= P(Z≤-1.33)

As you see there is no need to know the actual value of the population mean since you have to subtract it when calculating the Z values.

P(Z≤1.33) - P(Z≤-1.33)= 0.90824 - 0.09176= 0.81648

I hope it helps!

Answer 2

Answer:

The probability is 0.789

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.

In this problem, we have that:

$\sigma = 16, n = 400, s = \frac{16}{\sqrt{400}} = 0.8$

Find the probability that the mean age of the individuals sampled is within one year of the mean age for all individuals in the city.​

Using $\mu = 30$, this is the pvalue of Z when X = 31 subtracted by the pvalue of Z when X = 29. So

X = 31

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

By the Central Limit Theorem

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

$Z = \frac{31 - 30}{0.8}$

$Z = 1.25$

$Z = 1.25$ has a pvalue of 0.8944

X = 29

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

$Z = \frac{29 - 30}{0.8}$

$Z = -1.25$

$Z = -1.25$ has a pvalue of 0.1056

0.8944 - 0.1056 = 0.7888

Rounded to three decimal places, 0.789

The probability is 0.789