How to Calculate Mean and Median for Scaled Examination Marks?

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Homework Statement



The pass mark of an examination is set at 40. The following table shows some summary statistics of the marks (x) of the 200 students who took the examination.

Mark Range: x >= 40, x < 40
Number of students: 160, 40
Mean: 64.0, 32.0
Standard deviation: 6.0, 4.0

The teacher wants to scale the marks so that more students will pass the examination. The new mark corresponding to x is

y = 50 + \frac{5}{6}(x - 40), if x >= 40
y = \frac{5x}{4} if x < 40

(a) Find the mean and standard deviation of the new marks of the 200 students.
(b) The median of the original 200 marks is 52. Find the median of the new marks.

(Answer:
(a) 64; 13
(b) 60)

Homework Equations



Basic Formulae for Statistical Measures

The Attempt at a Solution



I don't know how to find the standard deviation for part (a) and the median for part (b).

Can anyone tell me how to solve them?

Thank you very much!
 
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I know that part (b) can be solved using linear transformation y = ax + b.

50 + \frac{5}{6}(52 - 40) = 60 but not \frac{5(52)}{4}.

Can anyone tell me the reason?

Thank you very much!
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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