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

Answer:

P(X<1200) is 0.8212

Step-by-step explanation:

Test statistic (z) = (X - mean)/sd

X is score of a tester = 1200

mean = 1497

sd = 322

z = (1200 - 1497)/322 = -297/322 = -0.92

The cumulative area of the test statistic is the probability that X<1200. The cumulative area is 0.8212.

Therefore, P(X<1200) = 0.8212

Answer 2

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.