Question

The distribution of SAT scores of all college-bound seniors taking the SAT in 2014 was approximately normal with a mean of 149714971497 and standard deviation of 322322322. Let XXX represent the score of a randomly selected tester from this group. Find P(X>1800)P(X>1800)P, (, X, is greater than, 1800, ).

Answer 1

The probability that a randomly selected person scored above 1800 on the SAT is approximately 17.36%, after calculating the corresponding z-score and looking up the probability in the Standard Normal Distribution table.

Step-by-step explanation:

To find P(X>1800), we first need to calculate the z-score for an SAT score of 1800. The z-score is computed as:

z = (X - μ) / σ

Where X is the SAT score, μ is the mean, and σ is the standard deviation. Given μ = 1497 and σ = 322, we have:

z = (1800 - 1497) / 322 = 303 / 322 ≈ 0.941

Once we have the z-score, we can use the Standard Normal Distribution table to find P(Z > 0.941). We find that P(Z > 0.941) ≈ 0.1736. Thus, the probability that a randomly selected college-bound senior has an SAT score above 1800 is approximately 0.1736 or 17.36%.

Answer 2

P ( X > 1800) = 0.1734

Step-by-step explanation:

Given:-

- The mean, u = 1497

- The standard deviation, s.d = 322

Find:-

P(X>1800)

Solution:-

- We will denote a random variable X that follows a normal distribution for the SAT scores in 2014 with parameters mean (u) and standard deviation (s.d) as follows:

X ~ N ( 1497 , 322 )

- The following probability can be calculated by first computing the Z-score value:

P ( X < x ) = P ( X < Z )

Where,

Z = ( x - u ) / s.d

- P(X > 1800) have the corresponding Z-score value:

Z = ( 1800 - 1497 ) / 322

Z = 0.941

- Hence, using Z-table:

P ( X > 1800) = 1 - P ( Z < 0.9471 )

P ( X > 1800) = 1 - 0.8266

P ( X > 1800) = 0.1734