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Answer :
Final answer:
a) The proportion of Puerto Rico high school students in the class of 2019 who had SAT total scores above 1400 is approximately 0.19%.
b) The 90th percentile of the SAT total scores of Puerto Rico high school students in the class of 2019 is approximately 1160.8.
c) The probability that exactly one of the ten Puerto Rico high school students in the class of 2019 had a SAT total score above 1400 is approximately 1.87%.
Explanation:
To solve the given questions, we will use the concept of z-scores and the standard normal distribution table.
a) Find the proportion of Puerto Rico high school students in the class of 2019 who had SAT total scores above 1400:
To find the proportion, we need to calculate the z-score for 1400 using the formula:
z = (x - mean) / standard deviation
Substituting the given values:
z = (1400 - 944) / 180 = 2.89
Using the standard normal distribution table, we can find that the proportion of students with SAT total scores above 1400 is approximately 0.0019 or 0.19%.
b) Find the 90th percentile of the SAT total scores of Puerto Rico high school students in the class of 2019:
The 90th percentile represents the score below which 90% of the students fall. To find this, we need to find the z-score corresponding to the 90th percentile using the standard normal distribution table.
From the table, we find that the z-score corresponding to the 90th percentile is approximately 1.28.
Using the formula for z-score:
z = (x - mean) / standard deviation
Substituting the given values:
1.28 = (x - 944) / 180
Solving for x, we get:
x = 1.28 * 180 + 944 = 1160.8
Therefore, the 90th percentile of the SAT total scores is approximately 1160.8.
c) For a random sample of ten Puerto Rico high school students in the class of 2019, find the probability that exactly one of them had a SAT total score above 1400:
To find the probability, we need to use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
where n is the sample size, k is the number of successes, and p is the probability of success.
In this case, n = 10, k = 1, and p = 0.0019 (proportion from part a).
Substituting the values into the formula:
P(X = 1) = (10 choose 1) * (0.0019)^1 * (1 - 0.0019)^(10 - 1)
Simplifying the expression:
P(X = 1) = 10 * 0.0019 * 0.9981^9
Calculating the probability, we find that the probability that exactly one of the ten students had a SAT total score above 1400 is approximately 0.0187 or 1.87%.
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