Question

SAT scores (out of 2400) are distributed normally with a mean of 1500 and a standard deviation of 300. Suppose a school council awards a certificate of excellence to all students who score at least 1900 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 2200

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

$\mu = 1500, \sigma = 300$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1900 - 1500}{300}$

$Z = 1.33$

$Z = 1.33$ 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2200 - 1500}{300}$

$Z = 2.33$

$Z = 2.33$ has a pvalue of 0.9901.

So $P(A \cap B) = 1 - 0.9901 = 0.0099$

Then

$P(B|A) = \frac{0.0099}{0.0918} = 0.1078$

10.78% probability this student's score will be at least 2200