Scores on a test follow normal distribution with mean of 460 and SD of 100. If all students in class of 41 attend the test what is probability that the given class will obtain a mean score of above 589.93? I can work out the Z score but am confused by the second part of the question. Do i need to work out T-score too ?
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I'm confused by the question too. The probability it seems to describe is so extremely small that I wonder why someone would even ask what it is. – David K Mar 30 '15 at 12:54
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The distribution of mean scores follows a $N(460, \frac{100}{41})$ distribution. Convert $589.93$ to a $Z$ score with respect to this distribution (call this z-score $z=\frac{589.93-460}{\sqrt{100/41}}$), and then calculate the probability a standard normal is above this z: $P(Z \geq z) = 1 - P(Z \leq z)=1- \Phi(\frac{589.93-460}{\sqrt{100/41}})$.
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