Suppose you have discovered a new species of walrus and are very excited aboutit. After a bit of research, you realize that the average weight of this species ofwalrus (in lbs) is approximately normal with meanμ= 3500 and varianceσ2= 75.What is the probability that you observe a walrus that weights more than 3520 lbs?

Respuesta :

Answer:

[tex]P(X>3520)=P(\frac{X-\mu}{\sigma}>\frac{3520-\mu}{\sigma})=P(Z>\frac{3520-3500}{8.66})=P(z>2.309)[/tex]

And we can find this probability using the complement rule:

[tex]P(z>2.309)=1-P(z<2.309)=1-0.990=0.010 [/tex]

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". Ā 

Solution to the problem

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

[tex]X \sim N(3500,\sigma=\sqrt{75}=8.66)[/tex] Ā 

Where [tex]\mu=3500[/tex] and [tex]\sigma=8.66[/tex]

We are interested on this probability

[tex]P(X>3520)[/tex]

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

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(X>3520)=P(\frac{X-\mu}{\sigma}>\frac{3520-\mu}{\sigma})=P(Z>\frac{3520-3500}{8.66})=P(z>2.309)[/tex]

And we can find this probability using the complement rule:

[tex]P(z>2.309)=1-P(z<2.309)=1-0.990=0.010 [/tex]