SAT scores (out of 1600) are distributed normally with a mean of 1100 and a standard deviation of 200. Suppose a school council awards a certificate of excellence to all students who score at least 1350 on the SAT, and suppose we pick one of the recognized students at random. What is the probability this student’s score will be at least 1500?

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.

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