Respuesta :
Answer:
10.78% probability this student's score will be at least 2200
Step-by-step explanation:
Normal distribution:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [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.
Conditional probability:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
[tex]\mu = 1500, \sigma = 300[/tex]
We pick a recognized student. What is the probability this student's score will be at least 2200
Event A: Recognized(scored at least 1900).
Event B: At least 1900.
Probability of scoring at least 1900.
1 subtracted by the pvalue of Z when X = 1900. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1900 - 1500}{300}[/tex]
[tex]Z = 1.33[/tex]
[tex]Z = 1.33[/tex] has a pvalue of 0.9082.
1 - 0.9082 = 0.0918.
So P(A) = 0.0918
Intersection:
The intersection between at least 1900 and at least 2200 is at least 2200.
Probability:
1 subtracted by the pvalue of Z when X = 2200. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{2200 - 1500}{300}[/tex]
[tex]Z = 2.33[/tex]
[tex]Z = 2.33[/tex] has a pvalue of 0.9901.
So [tex]P(A \cap B) = 1 - 0.9901 = 0.0099[/tex]
Then
[tex]P(B|A) = \frac{0.0099}{0.0918} = 0.1078[/tex]
10.78% probability this student's score will be at least 2200
