# Basic Statistical Measures

1. Mar 15, 2009

### chrisyuen

1. The problem statement, all variables and given/known data

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.

(a) 64; 13
(b) 60)

2. Relevant equations

Basic Formulae for Statistical Measures

3. 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!

Last edited: Mar 16, 2009
2. Mar 16, 2009

### chrisyuen

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!

Last edited: Mar 16, 2009