Question

Over the past decade, the mean number of hacking attacks experienced by members of the Information Systems Security Association is 510 per year with a standard deviation of 14.28 attacks. The distribution of number of attacks per year is normally distributed. Suppose nothing in this environment changes.
1. What is the likelihood that this group will suffer an average of more than 600 attacks in the next 10 years?

Answer 1

Answer:

$P(X>600)=P(\frac{X-\mu}{\sigma}>\frac{600-\mu}{\sigma})=P(Z>\frac{600-510}{14.28})=P(z>6.302)$

And we can find this probability using the complement rule and the normal standard distribution and we got:

$P(z>6.302)=1-P(z$

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

$X \sim N(510,14.28)$  

Where $\mu=510$ and $\sigma=14.28$

We are interested on this probability

$P(X>600)$

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:

$P(X>600)=P(\frac{X-\mu}{\sigma}>\frac{600-\mu}{\sigma})=P(Z>\frac{600-510}{14.28})=P(z>6.302)$

And we can find this probability using the complement rule and the normal standard distribution and we got:

$P(z>6.302)=1-P(z$