The percentage of scores that are less than 144 is 85.31%, given the normal distribution.
In the question, we are given that scores on a test are normally distributed with a mean of 123 and a standard deviation of 20.
We are asked to find the percent of scores that are less than 144.
Thus, we have:
Mean, μ = 123,
Standard Deviation, σ = 20.
The value to check, x = 144.
To find the percentage score, we will use the Z-Score table, with the formula, z = (x - μ)/σ.
We are asked to calculate the percentage of scores less than 144, that is,
P(X < 144)
= P(Z < (144 - 123)/20)
= P(Z < 21/20)
= P(Z < 1.05)
= 0.8531 {From the table}
= 85.31%.
Thus, The percentage of scores that are less than 144 is 85.31%, given the normal distribution.
Learn more about the normal distribution at
https://brainly.com/question/4079902
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