Junior Cert Maths Samples

_________________________________________ juniorcertsolutions.com ________________________________________ 2016 Question 3 – Higher Paper 2 Solution (a) Lower quartile Median Upper quartile 2015 19.5% 26% 36.5% The lower quartile was found as follows: 28 1 29 7.25 4 4 + = = th number or sum of 7th 8th 19 20 19.5% 2 2 + + = = The median is 28 1 th 2 + number = 14.5th: 26 26 26% 2 + = The upper quartile is 28 1 3 21.75 4 +   × =     : 36 37 36.5% 2 + = (b) The value of the lower quartile. This is the median of the lower half of the data or the middle of the lower half of the data (c) Mid interval 0 9 4.5 2 + = 14.5 24.5 34.5 44.5 % of members 0–9 10–19 20–29 30–39 40–49 2005 2 10 8 7 1 2015 0 7 10 8 3 (d) Mean for 2005: ( ) ( ) ( ) ( ) ( ) 2 4.5 14.5 10 24.5 8 34.5 7 44.5 1 636 22.71 22.7% 28 28 + + + + = = = (e) (i) Tom’s estimate for the mean of 2015 is 27% Actual mean is 26.86% Difference is due to the fact that Tom used mid-intervals (ii) Error is 27 26.86% 0.14% − = % error is 0.14 100 0.52% 26.86 1 × = (f) To show the data, draw two histograms 9 19 29 39 40 9 19 29 39 40