ohgodamnitw0

2022-09-03

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 ?

Brendan Bradley

Beginner2022-09-04Added 11 answers

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:

$P(Z\ge z)=1-P(Z\le z)=1-\mathrm{\Phi}(\frac{589.93-460}{\sqrt{100/41}})$

$P(Z\ge z)=1-P(Z\le z)=1-\mathrm{\Phi}(\frac{589.93-460}{\sqrt{100/41}})$

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