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Bsb123 Tha 2

In: Business and Management

Submitted By dan1el3C
Words 753
Pages 4
Euy Hyun Chong
N9718605
Euy Hyun Chong
N9718605
Numerical descriptive measures
Take Home Assignment - Topic 2
Numerical descriptive measures
Take Home Assignment - Topic 2

Question 1:

Data Set | | | | | | | | | 11 | 12 | 14 | 15 | 16 | 18 | 24 | 25 | 28 | 32 | Min | Quartile 1 | | Median | | Quartile 3 | Max |
Table [ 1 ]. Ages of 10 randomly selected students in a Judo school in Brisbane. (a) Mean (x) : 19.5
Standard Deviation (S) : 7.24568

(b) Median: 16+182=17

When comparing the mean and the median it tells the distribution is skewed to the right, which is positively skewed. The skewness of the dataset is positive due to bigger mean value than the median, as mean pulls the dataset to the right.

(c) Coefficient of Variation: CV= Sx= 7.2519.5=0.37157

(d) Using Excel: * First Quartile (Q1) = 13.5 [=QUARTILE.EXC(array,quart)] * Third Quartile (Q3) = 25.75 [=QUARTILE.EXC(array,quart)] * 80th Percentile = 27.4 [=PERCENTILE.EXC(array,k)]
Working out by common sense: * First Quartile (Q1) = 12+142=13 * Third Quartile (Q3) = 25+282=26.5 * 80th Percentile = Lp=n+1×P100=10+1×80100=8.8 ≈9th value * Therefore, the approximate of 80th percentile is 28 (9th value in data set).

(e) Range: highest value – lowest value = 32 – 11 = 21
IQR: Q3-Q1=12.25 (Used excel values)

(f) Data Set | | | | | | | | | 11 | 12 | 14 | 15 | 16 | 18 | 24 | 25 | 28 | 82 | Min | Quartile 1 | | Median | | Quartile 3 | Max |
Table [ 2 ]. Ages of 10 randomly selected students in a Judo school in Brisbane (Corrected data set)
Mean: 11+12+14+15+16+18+24+25+28+8210=24.5
Median: 16+182 =17
Range: 82 – 11 = 71
IQR: 12.25

The correct age 82, is relatively a large number in the...

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