Question

Students provided feedback to Dr. Soesmanto stating how the Online Quiz 3 in 1305AFE is substantially more challenging than the Online Quiz 1 or the Online Quiz 2. In examining this claim, Dr. Soesmanto took a random sample of 69 students and recorded their marks for Online Quiz 3. After collecting these data, he used Excel to run the numerical descriptive analysis and produced the following outputs.

Marks

Mean 5.347826 Standard Error 0.279924 Median 6 Mode 7

Standard

Deviation 2.32522 Sample Variance 5.40665 Kurtosis -0.83175 Skewness 0.004167 Range 7 Minimum 2 Maximum 9 Sum 369 Count 69 64th Percentile 6.8

Assist Dr. Soesmanto in interpreting the Excel report by answering these questions.

a) Provide an interpretation of the reported value of the mean in the context of the

problem. 1 mark

b) Provide an interpretation of the reported value of the median in the context of the

problem. 1 mark

c) Provide an interpretation of the reported value of the 64th percentile in the context of

the problem. 1 mark

d) If you were to draw the histogram of the data set, what shape of distribution do you think the data collected would resemble? Provide a brief justification to your answer. 1.5 marks

e) Calculate the approximate standard deviation using the relevant formula. Display working. Provideareasononwhytheapproximatestandarddeviationyoucalculated differs substantially to the standard deviation value reported in the table.

1.5 marks

Answer 1

a) Mean=5.347826

Interpretation:

Mean is a measure of central tendency that gives us an idea about where the centre of the data is located . From the mean we can tell that the average marks of students of Online Quiz 3 is 5.3478 that indicates that marks of students in Online Quiz 3 will be around 5.3478 on average.

b) Median=6

Interpretation:

Just like mean Median also give idea where the centre of the data is located .However, unlike Mean, Median takes into account of the position. Since Median=6, we can tell the middlemost marks of students of Online Quiz 3 is 6 i.e 50% of the marks will fall below and above 6.

c) 64th percentile=6.8.

Interpretation:

64th percentile is 6.8. So it means that 64% of the students got marks less than or equal to 6.8.

d) Here Mean=5.347826 ,Median=6 and Mode=7

So , Mean < Median < Mode

That indicates the distribution of marks is negatively skewed.

So if we draw a histogram the shape will be skewed to the left.

e) Approximate standard deviation

is s~ (Max-Min)/4 i.e Range/4=7/4=1.75

The standard deviation in the report is calculated using the proper formula of standard deviation as follows:

√(∑(xi-mean)^2/(n-1))

It involves the sum of squared of the deviation from mean but the approximate formula doesn't do that  and considers the range instead. So they give different estimates.