Question

The distribution of SAT scores of all college-bound seniors taking the SAT in 2014 was approximately normal with a mean of 14971497 1497 1497 and standard deviation of 322 322 322 322 . Let X=X= X= X, equals the score of a randomly selected tester from this group. Find P(X<1200)P(X<1200) P(X<1200) P, (, X, is less than, 1200, ).

Answer 1

Given Information:

Mean = μ = 1497

Standard deviation = σ = 322

test value = x = 1200

Required Information:

P(x < 1200) = ?

Answer:

P(x < 1200) = 0.17879

Explanation:

First we will find the z-score

P(x < X) = P(z < (x - μ)/σ)

P(x < 1200) = P(z < (1200 - 1497)/322)

P(x < 1200) = P(z < -0.92)

The z-score corresponding to z < -0.92 from z-table is given by

P(z < -0.92) = 0.17879

Therefore, the probability that SAT the test score will be less than 1200 is 0.17879.

Answer 2

Answer:

0.17, or 17%

(I got it right on Khan Academy)

Step-by-step explanation:

Mean = 1497

Standard Deviation = 322

1800 is 0.94 standard deviations away from the mean, and using a z-table, the value for 0.94 is 0.8264. But since it's 0.94 standard deviations above the mean, and 82.264% is much too large, we have to use 1 - x, where x = 0.8264. This leaves us with approximately 0.17, which makes sense and is the answer.

Hope this helps! Brainliest, please :)