Respuesta :
Answer:
0% probability of a random classmate getting a score more than 80
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation(which is the square root of the variance) [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 65, \sigma = \sqrt{12} = 3.4641[/tex]
What is the the probability of a random classmate getting a score more than 80?
This is 1 subtracted by the pvalue of Z when X = 80. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 65}{3.4641}[/tex]
[tex]Z = 4.33[/tex]
[tex]Z = 4.33[/tex] has a pvalue of 1
1 - 1 = 0
0% probability of a random classmate getting a score more than 80
