s1 revision / past paper questions / ch 1 - 4
DESCRIPTION
1. MATHEMATICAL MODELLING2. DATA LOCATION.3. MEASURES OF DISPERSION.4.REPRESENTATION OF DATA STEM LEAF DIAGRAMS INTERPOLATION METHOD BOX PLOTS HISTOGRAMSTRANSCRIPT
1. [Jan 2009 Q(4) ] Ans : median = 53 , LQ = 35, UQ = 60
2. [Jan 2009 Q(5) ] Ans : (a) w = 5 , h= 0.75 (b) 10.2 , IQR = 9 , (c) 12.8 , Sd = 9.88 (d) Median < mean, So positive skew (e) Use median and IQR, since data is skewed or not affected by extreme values or outliers
PAST PAPER QUESTIONS
S1 – Data Location, Measures of dispersion and Representation of data
3. [ June 2009 Q(3) ] Ans : (a) 1 cm (b) 6 cm
4. [ June 2009 Q(4) ] Ans : (a) 17.1 (b) mean = 17.1 , Sd = 3.28 (c) -0.00802 No skew/slightly skew.
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5. [ Jan 2010 Q(2) ] Ans : (a) 33 (b) IQR = 16 (c) Q1−IQR=24−16=8 , so7 is only ou tlier
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6. [ Jan 2010 Q(3) ] Ans : (a) 2.75 (b) 3.2273 (c) 0.421….(d) 3.25(e) mean< median , so negative skewness
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7. [ June 2010 Q(5)] Ans : (a) 23 , 35.5 (b) w= 2, h= 10.4 (c) mean= 23.5 , sd=10.7 (d) 23.7 or 23.9(e)
8. [ Jan 2011 Q(2) ] Ans : (a) 3.8 (b) It will have no effect since one is 4.5 under what it should be and the other is 4.5 above what it should be.
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9. [ Jan 2011 Q(3) ] Ans : (a) Outlier=25 , (b) negatively skewed since Q3−Q3 ¿Q2−Q1 (c) not true since the lower quartile is 7000 and therefore 75% above 7000 not 10000
10. [ Jan 2011 Q(5) ] Ans : (a) 41.3 (b) 44.1875 , 14.7 (c) mean > median therefore positive skew
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(5)
11. [ June 2011 Q(5) ] Ans : (a) 10.5 (b) 15.9 (c) 15.9 and 5.78 (d) Q2−Q1 ¿Q3−Q2
12. [ Jan 2012 Q(4) ] Ans : (a) 60 (b) 46,56,64 (c) 55.48 , 10.342.. (d) negative skewness, Mean < median < mode or Q2 – Q1 > Q3 – Q2 (e) mean=45, sd= 9
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13. [ June 2012 Q(5) ] Ans : (a) 90 (b) 28.8 (c) 28.1 (d) Q2<x , so positive skewness
(e) Since the data is skewed, median not affected by extreme values
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14. [ Jan 2013 Q(5) ] Ans : (a) w =4 , h= 2.25 cm, (b) 276 (c) 316 and 157 (d) 0.75 or 0.76 , positive skewness
15. [Jan 2007 Q(6) ]
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