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Date November 2021 Marks available 2 Reference code 21N.1.SL.TZ0.9
Level Standard Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number 9 Adapted from N/A

Question

A group of 120120 students sat a history exam. The cumulative frequency graph shows the scores obtained by the students.

The students were awarded a grade from 11 to 55, depending on the score obtained in the exam. The number of students receiving each grade is shown in the following table.

The mean grade for these students is 3.653.65.

Find the median of the scores obtained.

[1]
a.

Find an expression for aa in terms of bb.

[2]
b.

Find the number of students who obtained a grade 55.

[3]
c.i.

Find the minimum score needed to obtain a grade 55.

[2]
c.ii.

Markscheme

7575            A1

 

[1 mark]

a.

recognition that all entries add up to 120120          (M1)

a=120-6-13-26-ba=12061326b  OR  a=75-ba=75b            A1

 

[2 marks]

b.

6×1+13×2+26×3+(75-b)×4+b×5120=3.656×1+13×2+26×3+(75b)×4+b×5120=3.65              (M1)(A1)


Note: Award (M1) for attempt to substitute into mean formula, LHS expression is sufficient for the M mark. Award (A1) for correct substitutions in one variable OR in two variables, followed by evidence of solving simultaneously with a+b=75a+b=75.


(b=) 28(b=) 28             A1

 

[3 marks]

c.i.

120-120their part (c)(i) seen (e.g. 9292 indicated on graph)            (M1)

8484             A1

 

[2 marks]

c.ii.

Examiners report

Most candidates were able to identify the median from the cumulative frequency graph in part (a). In part (b), attention to the demand of this part of the question would have turned one mark into two for a significant number of candidates. Candidates realized that the sum of the frequencies in the table should add up to 120120, however, some neglected to make aa the subject of the formula.

In part (c)(i), there were mixed results when calculating the mean from the frequency table, with some candidates using only the frequencies and not including the grades in their calculation. Division by 55 was often seen. More able candidates, however, did manage to arrive at the required value of bb. In part (c)(ii) some candidates scored follow-through marks from their incorrect value for bb but it seemed that many candidates did not realize that they had to use the cumulative frequency graph to find the answer.

a.

Most candidates were able to identify the median from the cumulative frequency graph in part (a). In part (b), attention to the demand of this part of the question would have turned one mark into two for a significant number of candidates. Candidates realized that the sum of the frequencies in the table should add up to 120120, however, some neglected to make aa the subject of the formula.

In part (c)(i), there were mixed results when calculating the mean from the frequency table, with some candidates using only the frequencies and not including the grades in their calculation. Division by 55 was often seen. More able candidates, however, did manage to arrive at the required value of bb. In part (c)(ii) some candidates scored follow-through marks from their incorrect value for bb but it seemed that many candidates did not realize that they had to use the cumulative frequency graph to find the answer.

b.

Most candidates were able to identify the median from the cumulative frequency graph in part (a). In part (b), attention to the demand of this part of the question would have turned one mark into two for a significant number of candidates. Candidates realized that the sum of the frequencies in the table should add up to 120120, however, some neglected to make aa the subject of the formula.

In part (c)(i), there were mixed results when calculating the mean from the frequency table, with some candidates using only the frequencies and not including the grades in their calculation. Division by 55 was often seen. More able candidates, however, did manage to arrive at the required value of bb. In part (c)(ii) some candidates scored follow-through marks from their incorrect value for bb but it seemed that many candidates did not realize that they had to use the cumulative frequency graph to find the answer.

c.i.

Most candidates were able to identify the median from the cumulative frequency graph in part (a). In part (b), attention to the demand of this part of the question would have turned one mark into two for a significant number of candidates. Candidates realized that the sum of the frequencies in the table should add up to 120120, however, some neglected to make aa the subject of the formula.

In part (c)(i), there were mixed results when calculating the mean from the frequency table, with some candidates using only the frequencies and not including the grades in their calculation. Division by 55 was often seen. More able candidates, however, did manage to arrive at the required value of bb. In part (c)(ii) some candidates scored follow-through marks from their incorrect value for bb but it seemed that many candidates did not realize that they had to use the cumulative frequency graph to find the answer.

c.ii.

Syllabus sections

Topic 4—Statistics and probability » SL 4.2—Histograms, CF graphs, box plots
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Topic 4—Statistics and probability » SL 4.3—Mean, median, mode. Mean of grouped data, standard deviation. Quartiles, IQR
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