Based on data from a college, scores on a certain test are normally distributed with a mean of 1519 and a standard deviation of 316.
Find the percentage of scores less than 1298. round to two decimal places as needed
Given,
= 1519 , = 316
We convert this to standard normal as
P(X < x) = P(Z < ( x - ) / )
So,
P(X < 1298) = P(Z < ( 1298 - 1519 ) / 316 )
= P(Z < -0.70)
= 0.2422 (From Z table)
= 24.22%
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