Respuesta :
Answer:
0.2333 = 23.33% probability this student's score will be at least 2100.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution, and conditional probability.
Normal Probability Distribution:
Problems of normal distributions can be solved using 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
We use the conditional probability formula to solve this question. It is
[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.
SAT scores (out of 2400) are distributed normally with a mean of 1490 and a standard deviation of 295.
This means that [tex]\mu = 1490, \sigma = 295[/tex]
In this question:
Event A: Student was recognized.
Event B: Student scored at least 2100.
Probability of a student being recognized:
Probability of scoring at least 1900, which is 1 subtracted by the pvalue of Z when X = 1900. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1900 - 1490}{295}[/tex]
[tex]Z = 1.39[/tex]
[tex]Z = 1.39[/tex] has a pvalue of 0.9177
1 - 0.9177 = 0.0823
This means that [tex]P(A) = 0.0823[/tex]
Probability of a student being recognized and scoring at least 2100:
Intersection between at least 1900 and at least 2100 is at least 2100, so this is 1 subtracted by the pvalue of Z when X = 2100.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{2100 - 1490}{295}[/tex]
[tex]Z = 2.07[/tex]
[tex]Z = 2.07[/tex] has a pvalue of 0.9808
This means that [tex]P(A \cap B) = 1 - 0.9808 = 0.0192[/tex]
What is the probability this student's score will be at least 2100?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0192}{0.0823} = 0.2333[/tex]
0.2333 = 23.33% probability this student's score will be at least 2100.
