Find the total number of fish in the net.

Find (i) the modal length interval,

(ii) the interval containing the median length,

(iii) an estimate of the mean length.

(i) Write down an estimate for the standard deviation of the lengths.

(ii) How many fish (if any) have length greater than three standard deviations above the mean?

The fishing company must pay a fine if more than 10% of the catch have lengths less than 40cm.

Do a calculation to decide whether the company is fined.

A sample of 15 of the fish was weighed. The weight, W was plotted against length, L as shown below.

Exactly two of the following statements about the plot could be correct. Identify the two correct statements.

Note: You do not need to enter data in a GDC or to calculate r exactly.

(i) The value of r, the correlation coefficient, is approximately 0.871.

(ii) There is an exact linear relation between W and L.

(iii) The line of regression of W on L has equation W = 0.012L + 0.008 .

(iv) There is negative correlation between the length and weight.

(v) The value of r, the correlation coefficient, is approximately 0.998.

(vi) The line of regression of W on L has equation W = 63.5L + 16.5.

Total = 2 + 3 + 5 + 7 + 11 + 5 + 6 + 9 + 2 + 1     (M1)

(M1) is for a sum of frequencies.

= 51     (A1)(G2)

[2 marks]

Unit penalty (UP) is applicable where indicated in the left hand column.

(i) modal interval is 60 – 70

Award (A0) for 65     (A1)

(ii) median is length of fish no. 26,     (M1)(A1)

also 60 – 70     (G2)

Can award (A1)(ft) or (G2)(ft) for 65 if (A0) was awarded for 65 in part (i).

(iii) mean is \(\frac{{2 \times 25 + 3 \times 35 + 5 \times 45 + 7 \times 55 + ...}}{{51}}\)     (M1)

(UP) = 69.5 cm (3sf)     (A1)(ft)(G1)

Note: (M1) is for a sum of (frequencies multiplied by midpoint values) divided by candidate’s answer from part (a). Accept mid-points 25.5, 35.5 etc or 24.5, 34.5 etc, leading to answers 70.0 or 69.0 (3sf) respectively. Answers of 69.0, 69.5 or 70.0 (3sf) with no working can be awarded (G1).

[5 marks]

Unit penalty (UP) is applicable where indicated in the left hand column.

(UP) (i) standard deviation is 21.8 cm     (G1)

For any other answer without working, award (G0). If working is present then (G0)(AP) is possible.

(ii) \(69.5 + 3 \times 21.8 = 134.9 > 120\)     (M1)

no fish     (A1)(ft)(G1)

For ‘no fish’ without working, award (G1) regardless of answer to (c)(i). Follow through from (c)(i) only if method is shown.

[3 marks]

5 fish are less than 40 cm in length,     (M1)

Award (M1) for any of \(\frac{5}{51}\), \(\frac{46}{51}\), 0.098 or 9.8%, 0.902, 90.2% or 5.1 seen.

hence no fine.     (A1)(ft)

Note: There is no G mark here and (M0)(A1) is never allowed. The follow-through is from answer in part (a).

[2 marks]

(i) and (iii) are correct.     (A1)(A1)

[2 marks]

a) b), c) There was much confusion about how to present the intervals. Often the mid-point only was seen. (eg. 65 instead of 60-70). Understanding of mode, median and mean was usually good but too many candidates wasted time calculating standard deviation by hand and got it wrong. In c(ii) 'greater than three' caused no problems but 'above the mean' was often ignored.

a) b), c) There was much confusion about how to present the intervals. Often the mid-point only was seen. (eg. 65 instead of 60-70). Understanding of mode, median and mean was usually good but too many candidates wasted time calculating standard deviation by hand and got it wrong. In c(ii) 'greater than three' caused no problems but 'above the mean' was often ignored.

a) b), c) There was much confusion about how to present the intervals. Often the mid-point only was seen. (eg. 65 instead of 60-70). Understanding of mode, median and mean was usually good but too many candidates wasted time calculating standard deviation by hand and got it wrong. In c(ii) 'greater than three' caused no problems but 'above the mean' was often ignored.

d) This was often well done, even if earlier parts were poorly done.

e) Rather mixed performance here. It was hard to identify any consistency in the errors made.

Too much time was spent on this question. It was only worth two marks and candidates ought to have realised that it relied on a general pictorial understanding of the concepts, possibly supplemented by a little elementary arithmetic only, to compare (iii) and (vi). With good understanding, many of the options could be ruled out in a few seconds.