Question

3.In 1995, the Educational Testing Service in Princeton, New Jersey (which administers SAT exam) re-centered the scores so that the overall mean would be approximately 1,000 in the combined math and verbal scores for a "large standardized group". In 1996, approximately 1.1 million college-bound high school students took the exam and registered a mean score of 1,013, with a standard deviation of 222. About 40 percent of these students’ scores were between 900 and 1,100. a)Based on this estimate, what is the probability that of 10 randomly selected students, lessthan four will be between 900 and 1,100?

Answer 1

Answer:

38.22% probability that of 10 randomly selected students, lessthan four will be between 900 and 1,100

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they score between 900 and 1,100, or they score outside this range. The probability of a student scoring between 900 and 1,100 is independent from other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

About 40 percent of these students’ scores were between 900 and 1,100.

This means that $p = 0.4$.

a)Based on this estimate, what is the probability that of 10 randomly selected students, lessthan four will be between 900 and 1,100?

This is $P(X < 4)$ when $n = 10$. So

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0060$

$P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403$

$P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209$

$P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0060 + 0.0403 + 0.1209 + 0.2150 = 0.3822$

38.22% probability that of 10 randomly selected students, lessthan four will be between 900 and 1,100