Respuesta :
Answer:
For the company A : [tex] P(X>33000)= P(Z> \frac{33000-27000}{3000}) =P(Z>2) = 1-P(Z<2)= 0.0228[/tex]
For the company B: [tex] P(X>33000)= P(Z> \frac{33000-27000}{7000}) =P(Z>0.857) = 1-P(Z<0.857)= 0.196[/tex]
So as we can see we have a higher probability for the company B.
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".
Company A
Let X the random variable that represent the salaries of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(27000,3000)[/tex]
Where [tex]\mu=27000[/tex] and [tex]\sigma=3000[/tex]
For this case if we find the probability for [tex] P(X>33000)[/tex] using the z score given by:
[tex] z= \frac{x -\mu}{\sigma}[/tex]
And if we use this formula we got:
[tex] P(X>33000)= P(Z> \frac{33000-27000}{3000}) =P(Z>2) = 1-P(Z<2)= 0.0228[/tex]
Company B
Let X the random variable that represent the salaries of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(27000,7000)[/tex]
Where [tex]\mu=27000[/tex] and [tex]\sigma=7000[/tex]
For this case if we find the probability for [tex] P(X>33000)[/tex] using the z score given by:
[tex] z= \frac{x -\mu}{\sigma}[/tex]
And if we use this formula we got:
[tex] P(X>33000)= P(Z> \frac{33000-27000}{7000}) =P(Z>0.857) = 1-P(Z<0.857)= 0.196[/tex]
