Respuesta :
Answer:
[tex]P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-42}{20})=P(z>0.9)[/tex]
And we can find this probability with the complement rule and with the normal standard table or excel and we got:
[tex]P(z>0.9)=1-P(z<0.9)=1-0.816=0.184[/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(42,20)[/tex]
Where [tex]\mu=42[/tex] and [tex]\sigma=20[/tex]
We are interested on this probability
[tex]P(X>60)[/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>60)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-42}{20})=P(z>0.9)[/tex]
And we can find this probability with the complement rule and with the normal standard table or excel and we got:
[tex]P(z>0.9)=1-P(z<0.9)=1-0.816=0.184[/tex]
The probability that a randomly selected score would have a score greater than 60 points is 18.41%.
What is z score?
Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean) / standard deviation
Given; mean of 42 and a standard deviation of 20
For > 60 points:
z = (60 - 42)/20 = 0.9
P(z > 0.9) = 1 - P(z < 0.9) = 1 - 0.8159 = 0.1841
The probability that a randomly selected score would have a score greater than 60 points is 18.41%.
Find out more on z score at: https://brainly.com/question/25638875
