What Is the Value of k in This Normal Distribution Problem?

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The problem involves a normally distributed test score with a mean of 54 and a standard deviation of 9. To find the value of k, it is necessary to calculate the z-score for a score of 80, which is done using the formula z = (x - mean) / standard deviation. The number of students scoring 80 or better is derived from the cumulative probability, and this value is then doubled to find the number of students scoring k or less. By using the normal distribution table, the corresponding z-score can be determined, allowing for the calculation of k. The solution was successfully reached with this method.
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i am having trouble with this question...

On a test, the marks were normally distributed with a mean mark of 54 and a standard deviation of 9. If the number of students who received a mark less than or equal to k was twice the number of those who received a mark of 80 or better, then the value of k, correct to the nearest whole number, was...
 
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What have you done? You can calculate the "standard" variable corresponding to x= 0: z= (80- 54)/9 and look up the probability in a table. (There's probably one in your textbook but http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/normaltable.html
has one on the internet.) That's not actually the "number of students who received a mark of 80 or better"- it is actually "1- (number of student who received a mark of 80 or less/total number of students)". But subtracting it from 1 gives "number of students who received a mark of 80 or better/total number of students" and doubling that will then give "number of students who received a mark of k or less/total number of students" and then use that with the table to determine the corresponding "z" and finally solve for "x" (which, to the nearest integer, is your "k").
 
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i figured it out...thanks...
 
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