Question 928139
Ho: u = 16 
Ha: u >16 (claim)
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sample of 50: xbar =16.3 o= 1.2minutes. 
t(16.3) = .3/(1.2/sqrt(50)) = 1.7678
p-value = tcdf (1.7678, 100, 49) = .0417
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If Using z-value: normalcdf(1.7678, 100) = .0385 (Same Results)
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confidence level .05
.04 < .05   0r .0385 < .05
Reject Ho.  Test results support the claim u >16 
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re TY
mean of 1400 and a standard deviation of 200. 
A college will only take applicants in the top 2.5% or higher. 
What is the lowest score they would need on the entrance exam
200(invNorm(.975) + 1400 = 200(1.96) + 1400 = 1792, lowest score