Question

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval (Express the result to four significant digits.)

Answer 1

Answer:

$z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5$

$z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5$

And using a calculator, excel ir the normal standard table we have that:

$P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)$

And we can calculate the probability like this:$P(-1.5 \leq Z \leq 1.5) = P(z$Step-by-step explanation:

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47<=X<53. Is the assumption of normality important. Why?

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(50,12)$  

Where $\mu=50$ and $\sigma=12$

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We can find the probability required like this:

$z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5$

$z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5$

And using a calculator, excel ir the normal standard table we have that:

$P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)$

And we can calculate the probability like this:

$P(-1.5 \leq Z \leq 1.5) = P(z$