Question

What is the probability that a randomly selected member of a normally distributed population will lie more than 1.8 standard deviations from the mean?Group of answer choices0.18410.03590.81590.0719

Answer 1

Answer:

$P(X> \mu +1.8\sigma)=P(\frac{X-\mu}{\sigma}>\frac{\mu +1.8\sigma-\mu}{\sigma})=P(Z>1.8)$

And we can find this probability using the complement rule:

$P(z>1.8)=1-P(z$

And using the normal standard distirbution table or excel we got:

$P(z>1.8)=1-P(z$

Step-by-step explanation:

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 if interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu,\sigma)$  

We are interested on this probability

$P(X>\mu +1.8 \sigma)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X> \mu +1.8\sigma)=P(\frac{X-\mu}{\sigma}>\frac{\mu +1.8\sigma-\mu}{\sigma})=P(Z>1.8)$

And we can find this probability using the complement rule:

$P(z>1.8)=1-P(z$

And using the normal standard distirbution table or excel we got:

$P(z>1.8)=1-P(z$