Question 926144
At a certain university the cumulative grade point average (CGPA) of first year students usually averages 2.48 with a standard deviation 0.42 . It has been found that the marks are usually approximately normally distributed. 
a) What is the probability that a student will have a CGPA that is between 1 and 3 ?
z(1) = (1-2.48)/0.42 =  -3.5238
z(3) = (3-2.48)/0.42 =  1.2381
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P(1< x < 3) = P(-3.5238 < z < 1.2381) = normalcdf(-3.5338,1.2381) = 0.8920
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 b) What percent of students will be on probation, i.e. their CGPA is less than 2 ?
z(2) = (2-2.48)/0.42 = -1.1429
P(x < 2) = P(z < -1.1429) = normalcdf(-100,-1.1429) = 0.1265
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 c) Academic scholarships are awarded to the top 5% of first year students. What minimum CGPA is needed to receive a scholarship?
Find the z-value with a right-tail of 0.05
invnorm(0.95) = 1.645
Find the corresponding score::
x = 1.645*0.42 + 2.48 = 3.1708
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Cheers,
Stan H.
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