Suppose SAT Writing scores are normally distributed with a mean of 497 and a standard deviation of 109. A university plans to aw
ard scholarships to students whose scores are in the top 2%. What is the minimum score required for the scholarship? Round your answer to the nearest whole number, if necessary.
1 answer:
Answer:
721.54
Step-by-step explanation:
We have to convert the 2% given in the statement into a z-score, as follows:
P (X> x) = 2% = 0.02, P (Z> z) = 0.02
thus find z such that:
P (Z <z) = 1 - P (Z> z)
P (Z <z) = 1 - 0.02
P (Z <z) = 0.98
we look for what value of z corresponds to in the normal distribution table and it is 2.06
x = m + z * sd
m is mean and sd standard deviation, replacing:
x = 497 + 2.06 * 109
x = 721.54
721.54 would be the minimum score.
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Actually Welcome to the Concept of the Average rate of change.
The average rate of change can be calculated by, finding the change of the two readings and getting a average between them.
here, for x =3 to x =5
1.) in the interval, x =3 to x = 4
the change is ==> 287 - 125 = 162
2.) in the interval, x = 4 to x=5
the change is ==> 549-287 = 262
Now the average ==> 162+262/2 ===> 212
Hence the average rate of change is 212
Answer: Choice A
Raising something to the 1/4th power is the same as applying the 4th root to that expression. Raising an expression to the 1/n power is the same as applying the nth root.
Rule:
![x^{1/n} = \sqrt[n]{x}](https://tex.z-dn.net/?f=x%5E%7B1%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%7D%20)
(if the text is too small it basically says x to the (1/n) = square root with a small 'n' all over x)
Raising something to the 1/4th power is the same as applying the 4th root to that expression. Raising an expression to the 1/n power is the same as applying the nth root.
Rule:
(if the text is too small it basically says x to the (1/n) = square root with a small 'n' all over x)