Students pass a test if they score 50% or more. The marks of a large number of students were sampled and the mean and standard deviation were calculated as 42% and 8% respectively. Assuming this data is normally distributed, what percentage of students pass the test?

Respuesta :

Answer:

[tex]P(X>50)=P(\frac{X-\mu}{\sigma}>\frac{50-\mu}{\sigma})=P(Z>\frac{50-42}{8})=P(z>1)[/tex]

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

[tex]P(z>1)=1-P(z<1)=1-0.841=0.159[/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".  

Solution to the problem

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

[tex]X \sim N(42,8)[/tex]  

Where [tex]\mu=42[/tex] and [tex]\sigma=8[/tex]

We are interested on this probability

[tex]P(X>50)[/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>50)=P(\frac{X-\mu}{\sigma}>\frac{50-\mu}{\sigma})=P(Z>\frac{50-42}{8})=P(z>1)[/tex]

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

[tex]P(z>1)=1-P(z<1)=1-0.841=0.159[/tex]