For large classes, test scores are usually approximately normally distributed. S
ID: 3179781 • Letter: F
Question
For large classes, test scores are usually approximately normally distributed. Suppose 500 students have taken a Statistics test. The distribution of scores is approximately Normal with mean 79.68 and standard deviation 17.91 (these happen to be the values from your first midterm exam). What proportion of students received an "A" (greaterthanorequalto 90 points) on the test? If a professor "curves" a class, that means that the professor uses the exam score distribution itself to assign grades. Where would the professor have set the A/B cutoff if she would have wanted to assign A's to the top 15% of the class?Explanation / Answer
Mean ( u ) =79.68
Standard Deviation ( sd )=17.91
Normal Distribution = Z= X- u / sd ~ N(0,1)
a.
P(X < 90) = (90-79.68)/17.91
= 10.32/17.91= 0.5762
= P ( Z <0.5762) From Standard Normal Table
= 0.7178
P(X > = 90) = (1 - P(X < 90)
= 1 - 0.7178 = 0.2822
28.22% are received greater than 90
b.
P ( Z > x ) = 0.15
Value of z to the cumulative probability of 0.15 from normal table is 1.0364
P( x-u/ (s.d) > x - 79.68/17.91) = 0.15
That is, ( x - 79.68/17.91) = 1.0364
--> x = 1.0364 * 17.91+79.68 = 98.2425
the cutoff score would be 98.2425
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