S1 January 2011 Q5

5. On a randomly chosen day, each of the 32 students in a class recorded the time, \(t\) minutes to the nearest minute, they spent on their homework. The data for the class is summarised in the following table.

Time, \(t\)	Number of students
10 – 19	2
20 – 29	4
30 – 39	8
40 – 49	11
50 – 69	5
70 – 79	2

(a) Use interpolation to estimate the value of the median. (2)

Given that

\[\sum t = 1414 \quad \text{and} \quad \sum t^2 = 69\,378\]

(b) find the mean and the standard deviation of the times spent by the students on their homework. (3)

(c) Comment on the skewness of the distribution of the times spent by the students on their homework. Give a reason for your answer. (2)

Mark scheme (a)

Scheme	Marks
Median = 32/2 = 16th term (16.5) \(\dfrac{x - 39.5}{49.5 - 39.5} = \dfrac{16 - 14}{25 - 14}\) or \(x = 39.5 + \left(\dfrac{2}{11} \times 10\right)\)	M1
Median = 41.3 ( use of \(n\) + 1 gives 41.8) (awrt 41.3)	A1
	(2)

Notes

M1 for an attempt to use interpolation to find the median. Condone use of 39 or 40 for 39.5
e.g. allow \(39 + \dfrac{2}{11} \times 10\) (o.e.) or \(40 + \dfrac{2}{11} \times 10\) (o.e.) to score M1A0 but must have the 10
A1 for awrt 41.3 (or awrt 41.8 if using (\(n\) + 1))

Mark scheme (b)

Scheme	Marks
Mean= \(\dfrac{1414}{32} = 44.1875\) (awrt 44.2)	B1
Standard deviation \(= \sqrt{\dfrac{69378}{32} - \left(\dfrac{1414}{32}\right)^2}\)	M1
= 14.7 (or \(s\) = 14.9)	A1
	(3)

Notes

B1 for awrt 44.2
M1 for a correct expression including square root. (Allow ft of their mean)
A1 for awrt 14.7 (If using \(s\) for awrt 14.9)

Mid-points: You may see \(\sum t = 1339 \to \bar{t} = 41.8\) and \(\sum t^2 = 62928 \to \sigma\) 14.7 or \(s\) = 14.9
this scores B0 for the mean but can score M1 for a correct st.dev expression and A1 for ans.

Correct answer only in (a) and (b) can score full marks but check (\(n\) +1) case in (a)

Mark scheme (c)

Scheme	Marks
mean > median therefore positive skew	B1ft B1ft
	(2)
	(7 marks)

Notes

1st B1ft for a correct comparison of their mean and their median (may be in a formula)
Calculating median – mean as negative is OK for this B1 but must say +ve skew for 2nd B1
Only allow comparison to be \(\approx 0\) if \(|\text{mean} - \text{median}| \leqslant 0.5\)

2nd B1ft for a correct description of skewness based on their values of mean and median.
ft their values for mean and median not their previous calculation/comparison
Must be compatible with their previous comparison (if they have one)
“Positive skew” with no reason is B0B1 provided you can see their values that imply that.
Description should be “positive” or “negative” or “no” skew or “symmetric”
“Positive correlation” is B0

Quartiles: 1st B1ft if \(Q_1\) = awrt 32 and \(Q_3\) = awrt 49 seen and a correct comparison made. ft \(Q_2\)
2nd B1ft if \(Q_1\) = awrt 32 or \(Q_3\) = awrt 49 seen and a correct description based on their quartiles and their comparison is made. (Should get “negative skew”)