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, ).

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

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step-by-step explanation:

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step-by-step explanation: