You were given the following data and asked to find the median and inter quartile range .

1784.8,1867.0,1954.9,2049.6,2060.4,1903.3,1828.6,1813.4,1773.4,1803.4

The median is found by using the Rule median number is the (n+1)/2 number , where n is the number of elements of data in this case the 5.5th number .

First we must write the data in order of magnitude 1773.4,1784.8,1803.4,1813.4,1828.6,1867.0,1903.3,1954.9,2049.6,2060.4

The median is (1828.6+1867.0)/2 =1847.8

**How do you find the interquartile rang ?**

Some suggest that the lower quartile is the middle number between the start and the median in this case 1803.4 and the upper quartile is the middle number between the median and the end in this case 1954.9 .

**Giving an interquartile range of 1954.9-1803.4= 151.5 .**

But using the n method the lower quartile is (n+1)/4=2.75th number. To find this we add .75 of the difference between 2nd and 3rd numbers to the 2nd number this gives 1784.8+.75(1803.4-1784.8)=**1798.75**

The upper quartile is 3(n+1)/4=8.25th number To find this we add .25 of the difference between the 8th and 9th number to the 8th number this gives 1954.9 +.25(2049.6-1954.9)=**1978.58**

the interquartile range by this method is **179.825**.

**Interquartile range calculators** : There are many interquartile calculators on the net.

**The Acula calculator** gives Lower quartile (x_{L}): 1798.75,Upper quartile (x_{U}): 1978.575,Interquartile range (x_{U}-x_{L}): 179.825 similar to the n method.

Another calculator gives the following result **Results**

** .25th Percentile: 1803.4**,.**50th Percentile: 1847.8**,.**75th Percentile: 1954.9** I**nterquartile Range: 151.5**

Notice all methods give the same median

From Wikipedia

**For discrete distributions, there is no universal agreement on selecting the quartile values. ^{[1]}**

- Use the median to divide the ordered data set into two halves. Do not include the median in either half.
- The lower quartile value is the median of the lower half of the data. The upper quartile value is the median of the upper half of the data.

This rule is employed by the TI-83 calculator boxplot and “1-Var Stats” functions.

- Use the median to divide the ordered data set into two halves. If the median is a datum (as opposed to being the mean of the middle two data), include the median in both halves.
- The lower quartile value is the median of the lower half of the data. The upper quartile value is the median of the upper half of the data.

- If there are an even number of data points, then the method is the same as above.
- If there are (4
*n*+1) data points, then the lower quartile is 25% of the*n*th data value plus 75% of the (*n*+1)th data value; the upper quartile is 75% of the (3*n*+1)th data point plus 25% of the (3*n*+2)th data point. - If there are (4
*n*+3) data points, then the lower quartile is 75% of the (*n*+1)th data value plus 25% of the (*n*+2)th data value; the upper quartile is 25% of the (3*n*+2)th data point plus 75% of the (3*n*+3)th data point.

This always gives the arithmetic mean of Methods 1 and 2; it ensures that the median value is given its correct weight, and thus quartile values change as smoothly as possible as additional data points are added.

Ordered Data Set: 6, 7, 15, 36, 39, 40, 41, 42, 43, 47, 49

Method 1 | Method 2 | Method 3 |
---|---|---|

It had to happen when you give every students sitting Higher maths who achieves a D3 (38% or better ) or 36% for students answering the exam through Irish) 25 extra points you would expect an increase in numbers sitting the exam. So it has come to pass in 2014 about have applied to sit **17,500** the H maths exam up from** 8390** in 2010, (14326 actually sat the exam) .Now most teachers would agree that the number of H maths students could not double in 4 years .Our PISA results would indicate only a modest improvement in standards. So this 100% increase in students taking H maths can only be attributed to the extra 25 points . This 100% increase in numbers since 2010 has led to problems in schools where teachers who have only taught ordinary level maths suddenly find themselves teaching a Leaving Higher class where the highest grade of the sydents at Junior cert was a C ! Schools that in the past had one/two leaving H classes now have 3 or 4 classes . All the grinds schools report huge numbers looking for classes in H maths for both 5th and 6th years.

**The reaction**

Faced with this problem and the fact that the failure rate for Higher maths has been as low 3.2% something had to give. This was compounded by the 2012 debacle. In 2012 the exam set was a full blown Project maths exam ,the initial results were a disaster ( some say a 10%+ failure rate) !Depending who you ask the exam was re corrected and marking scheme changed at least twice .Resulting in 55 marks out of 75 for part (a) of the robotic arm question and twenty marks out of 25 for writing down (1,1) for the circle question. **The failure rate in 2012 after many re corrections eventually ended up at 2.3%** !

The powers that be were determined this would not happen in 2013 .The 2013 paper was a conventional paper and included a question previously asked on the **1975 Inter cert Cert lower course**.

In 2014 faced with 17500 students at least 6,000 who should be doing Ord level the standards of the questions had to be lowered otherwise large numbers would fail .Which is exactly what has happened.

**Paper 1 (**includes comments from Mark Lynch and other teachers)

Q1.Probably LCOL standard.

Q2. Complex number, again LOL

Q3. Induction, Series, 2014 sample question asked the rest is Ordinary level .

Q4. Exactly the same as 1996/2001 old question 8 in LCOL paper but worth more marks.

Q5. Very basic integration question.

Q6. The most challenging of the first 6 questions. But is the same series as Q3 you could have used the results of Q3 here .

The so called problem solving questions ( well there were none )

Q7. This question was on **the JCH 2014** sample.

Q8. The nearest question to a proper LCH maths question this will be used to filter out the A and B students from the rest.

Q9. Decent question, only last part would have posed a challenge to our most able students again an A/B filter.

In comparison to the Pre Project maths this paper is significantly easier

2014 Leaving Cert H paper 2 analysis .

Q1 Trigonometry barely ordinary level standard .

Q2 Trigonometry very little work for 25 marks !

Q3 Probability same question asked on 2011 Q4 Junior Cert H sample .

Q4 Trigonometry Only leaving H Question so far and Physics students would have a distinct advantage is this fair?

Q5 Coordinate geometry of the line .This question has been asked before on LCH maths but was not worth 25 marks.

Q6A Geometry Junior cert H question (asked many times on the Junior cert in the past)

Q6B Geometry another Junior cert question would have been worth 15 marks on the Junior cert but worth 25 marks, asked previously on L C Ord 1998 and 2004 but worth only 20 mark.

Q7 Statistics A dreadful question barely Leaving Cert Ordinary level standard!

Q8 Probability About Ordinary level Leaving cert standard

Q9(a) Coordinate geometry of the circle no more difficult than Ordinary level question.

Q9(b)Geometry This exact question was asked before on the **1974 Inter cert H** .

**The following comments from Junior Minister Sean Sherlock may explain why standards have been allowed to fall.**

This is a reply received from the Junior Minister in relation to a question about falling standards in LCH Project maths.

“*Standards do not radically vary year on year for large cohorts of students. When new*

*knowledge and skill sets are being tested the standard required is difficult to tie down in*

*the first year.( We are now on our 4th year of exams in Project Maths!)*

*This is particularly the case with Project Maths where candidates are developing new skills*

*and more are sitting the exam at higher level. These candidates cannot be treated unfairly*

*compared to previous candidates. (What does this mean? Is this an admission that standards have been compromised?)*

*The exam papers and marking schemes for Project Maths are new. For this reason, some*

**latitude is required** in the development and implementation of the marking scheme. This is

*fully in line with international practice in large scale public exams like the Leaving*

Certificate” What evidence do we have of this?

Update :

**Leaving Cert project Maths Results (2014) V Pre Project maths results (2003-2009)**

Year | 2014 | 2003-2009 (average) | Comment |

Number of students | 14326 | 9081 | 57% increase in uptake |

A1 % | 3.9% (558) | 7.32 (664) | A1 rate down 45% |

A1+A2 | 10.2% (1461) | 14.9% (1353) | A1+A2 rate down 32% |

C or Better | 72.6% | 79.2% | C or better rate down 8% |

D1,D2,D3 | 22.9% | 17.5% | D rate up 31% |

E,F,G | 4.5% (655) | 3.7% (335) | Highest % failure rate in 14 years Largest number of students failing ever! |

100 points or more | 5616 | 655 | 9 times the Pre project number |

D3 (38% +) or 36% + if exam is done through Irish | 6.6% (966) got 70 points | 4.65 % (422) got 45 points | A D3 in 2014 = 70 points |

B3 (100 points) | 1619 | 655 | D3 pre project maths = 45 points |

Mean Mark | 64.32% | 67.5% | Mean mark down 3% |

Prove that the sum of the areas of the two shaded lunes is equal to the area of the triangle ABC.

Mark in the regions on the diagram as shown.

Solution From part (iii) We know that the areas A+B+C+D=B+C+E therefore A+D=E

All the solutions from all Leaving( H and Ord) and Junior (H) maths papers real and sample can be found at www.leavingcert.ie