Respuesta :
Answer:
0.2159 = 21.59% probability this student’s score will be at least 1500.
Step-by-step explanation:
To solve this question, we need to understand the normal 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.
In this question:
Event A: Recognized student(scored more than 1350)
Event B: Score of at least 1500.
SAT scores (out of 1600) are distributed normally with a mean of 1100 and a standard deviation of 200
This means that [tex]\mu = 1100, \sigma = 200[/tex]
Probability of being recognized.
1 subtracted by the pvalue of Z when X = 1350. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1350 - 1100}{200}[/tex]
[tex]Z = 1.25[/tex]
[tex]Z = 1.25[/tex] has a pvalue of 0.8944.
1 - 0.8944 = 0.1056
So [tex]P(A) = 0.1056[/tex]
Probabibility of being recognized and scoring at least 1500.
Intersection between more than 1350 and more than 1500 is more than 1500. So this probability is 1 subtracted by the pvalue of Z when X = 1500.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1500 - 1100}{200}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
1 - 0.9772 = 0.0228
So, [tex]P(A \cap B) = 0.0228[/tex]
What is the probability this student’s score will be at least 1500?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0228}{0.1056} = 0.2159[/tex]
0.2159 = 21.59% probability this student’s score will be at least 1500.
