Semester 1 Review Paper 2 2018 [106 marks]

1a.

A group of 800 students answered 40 questions on a category of their choice out of History, Science and Literature.

For each student the category and the number of correct answers, , was recorded. The results obtained are represented in thefollowing table.

State whether is a discrete or a continuous variable.

Markschemediscrete (A1)

[1 mark]

N

N

1b. Write down, for , the modal class;

Markscheme (A1)

[1 mark]

N

11 ⩽ N ⩽ 20

1c. Write down, for , the mid-interval value of the modal class.

Markscheme15.5 (A1)(ft)

Note: Follow through from part (b)(i).

[1 mark]

N

1d. Use your graphic display calculator to estimate the mean of ;

Markscheme (G2)

[2 marks]

N

21.2 (21.2125)

[1 mark]

[1 mark]

[1 mark]

[2 marks]

1e. Use your graphic display calculator to estimate the standard deviation of .

Markscheme (G1)

[1 marks]

N

9.60 (9.60428…)

1f.

A test at the 5% significance level is carried out on the results. The critical value for this test is 12.592.

Find the expected frequency of students choosing the Science category and obtaining 31 to 40 correct answers.

MarkschemeOR (M1)

Note: Award (M1) for correct substitution into expected frequency formula.

(A1)(G2)

[2 marks]

χ2

× × 800260800

157800

260×157800

= 51.0 (51.025)

1g. Write down the null hypothesis for this test;

Markschemechoice of category and number of correct answers are independent (A1)

Notes: Accept “no association” between (choice of) category and number of correct answers. Do not accept “not related” or “notcorrelated” or “influenced”.

[1 mark]

1h. Write down the number of degrees of freedom.

Markscheme6 (A1)

[1 mark]

1i. Write down the -value for the test;

Markscheme (G1)

[1 mark]

p

0.0644 (0.0644123…)

Write down the statistic.χ2

[1 mark]

[2 marks]

[1 mark]

[1 mark]

[1 mark]

1j.

Markscheme (G2)

[2 marks]

11.9 (11.8924…)

1k. State the result of the test. Give a reason for your answer.

Markschemethe null hypothesis is not rejected (the null hypothesis is accepted) (A1)(ft)

OR

(choice of) category and number of correct answers are independent (A1)(ft)

as OR (R1)

Notes: Award (R1) for a correct comparison of either their statistic to the critical value or their -value to the significance level.Award (A1)(ft) from that comparison.

Follow through from part (f). Do not award (A1)(ft)(R0).

[2 marks]

11.9 < 12.592 0.0644 > 0.05

χ2 χ2 p

2a.

A company performs an experiment on the efficiency of a liquid that is used to detect a nut allergy.

A group of 60 people took part in the experiment. In this group 26 are allergic to nuts. One person from the group is chosen at random.

Find the probability that this person is not allergic to nuts.

Markscheme (A1)(A1)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

( , 0.567, 0.566666… , 56.7%)3460

1730

2b.

A second person is chosen from the group.

Find the probability that both people chosen are not allergic to nuts.

[2 marks]

[2 marks]

[2 marks]

[2 marks]

Markscheme (M1)

Note: Award (M1) for their correct product.

(A1)(ft)(G2)

Note: Follow through from part (a).

[2 marks]

×3460

3359

= 0.317 ( , 0.316949… , 31.7%)187590

2c.

When the liquid is added to a person’s blood sample, it is expected to turn blue if the person is allergic to nuts and to turn red if theperson is not allergic to nuts.

The company claims that the probability that the test result is correct is 98% for people who are allergic to nuts and 95% for people whoare not allergic to nuts.

It is known that 6 in every 1000 adults are allergic to nuts.

This information can be represented in a tree diagram.

Copy and complete the tree diagram.

Markscheme

(A1)(A1)(A1)

Note: Award (A1) for each correct pair of branches.

[3 marks]

2d.

An adult, who was not part of the original group of 60, is chosen at random and tested using this liquid.

Find the probability that this adult is allergic to nuts and the liquid turns blue.

[3 marks]

[2 marks]

Markscheme (M1)

Note: Award (M1) for multiplying 0.006 by 0.98.

(A1)(G2)

[2 marks]

0.006 × 0.98

= 0.00588 ( , 0.588%)14725000

2e. Find the probability that the liquid turns blue.

Markscheme (A1)(ft)(M1)

Note: Award (A1)(ft) for their two correct products, (M1) for adding two products.

(A1)(ft)(G3)

Note: Follow through from parts (c) and (d).

[3 marks]

0.006 × 0.98 + 0.994 × 0.05 (0.00588 + 0.994 × 0.05)

= 0.0556 (0.05558, 5.56%, )277950000

2f. Find the probability that the tested adult is allergic to nuts given that the liquid turned blue.

Markscheme (M1)(M1)

Note: Award (M1) for their correct numerator, (M1) for their correct denominator.

(A1)(ft)(G3)

Note: Follow through from parts (d) and (e).

[3 marks]

0.006×0.980.05558

= 0.106 (0.105793… , 10.6%, )42397

2g.

The liquid is used in an office to identify employees who might be allergic to nuts. The liquid turned blue for 38 employees.

Estimate the number of employees, from this 38, who are allergic to nuts.

[3 marks]

[3 marks]

[2 marks]

Markscheme (M1)

Note: Award (M1) for multiplying 38 by their answer to part (f).

(A1)(ft)(G2)

Notes: Follow through from part (f). Use of 3 sf result from part (f) results in an answer of 4.03 (4.028).

[2 marks]

0.105793… × 38

= 4.02 (4.02015…)

3a.

The following table shows the average body weight, , and the average weight of the brain, , of seven species of mammal. Bothmeasured in kilograms (kg).

Find the range of the average body weights for these seven species of mammal.

Markscheme (M1)

(A1)(G2)

[2 marks]

x y

529 − 3

= 526 (kg)

3b. For the data from these seven species calculate , the Pearson’s product–moment correlation coefficient;

Markscheme (G2)

[2 marks]

r

0.922 (0.921857…)

3c. For the data from these seven species describe the correlation between the average body weight and the average weight of thebrain.

[2 marks]

[2 marks]

[2 marks]

Markscheme(very) strong, positive (A1)(ft)(A1)(ft)

Note: Follow through from part (b)(i).

[2 marks]

3d. Write down the equation of the regression line on , in the form .

Markscheme (A1)(A1)

Note: Award (A1) for , (A1) for 0.0923.

Award a maximum of (A1)(A0) if the answer is not an equation in the form .

[2 marks]

y x y = mx + c

y = 0.000986x + 0.0923 (y = 0.000985837… x + 0.0923391…)

0.000986x

y = mx + c

3e.

The average body weight of grey wolves is 36 kg.

Use your regression line to estimate the average weight of the brain of grey wolves.

Markscheme (M1)

Note: Award (M1) for substituting 36 into their equation.

(A1)(ft)(G2)

Note: Follow through from part (c). The final (A1) is awarded only if their answer is positive.

[2 marks]

0.000985837… (36) + 0.0923391…

0.128 (kg) (0.127829… (kg))

3f.

In fact, the average weight of the brain of grey wolves is 0.120 kg.

Find the percentage error in your estimate in part (d).

[2 marks]

[2 marks]

[2 marks]

Markscheme (M1)

Note: Award (M1) for their correct substitution into percentage error formula.

(A1)(ft)(G2)

Note: Follow through from part (d). Do not accept a negative answer.

[2 marks]

∣∣ ∣∣ × 1000.127829…−0.1200.120

6.52 (%) (6.52442... (%))

3g.

The average body weight of mice is 0.023 kg.

State whether it is valid to use the regression line to estimate the average weight of the brain of mice. Give a reason for youranswer.

MarkschemeNot valid (A1)

the mouse is smaller/lighter/weighs less than the cat (lightest mammal) (R1)

OR

as it would mean the mouse’s brain is heavier than the whole mouse (R1)

OR

0.023 kg is outside the given data range. (R1)

OR

Extrapolation (R1)

Note: Do not award (A1)(R0). Do not accept percentage error as a reason for validity.

[2 marks]

4a.

The table below shows the distribution of test grades for 50 IB students at Greendale School.

Calculate the mean test grade of the students;

Markscheme (M1)

Note: Award (M1) for correct substitution into mean formula.

(A1) (G2)

[2 marks]

=1(1)+3(2)+7(3)+13(4)+11(5)+10(6)+5(7)

5023050

= 4.6

[2 marks]

[2 marks]

4b. Calculate the standard deviation.

Markscheme (G1)

[1 mark]

1.46 (1.45602…)

4c. Find the median test grade of the students.

Markscheme5 (A1)

[1 mark]

4d. Find the interquartile range.

Markscheme (M1)

Note: Award (M1) for 6 and 4 seen.

(A1) (G2)

[2 marks]

6 − 4

= 2

4e.

A student is chosen at random from these 50 students.

Find the probability that this student scored a grade 5 or higher.

Markscheme (M1)

Note: Award (M1) for seen.

(A1) (G2)

[2 marks]

11+10+550

11 + 10 + 5

= ( , 0.52, 52%)2650

1325

4f.

A second student is chosen at random from these 50 students.

Given that the first student chosen at random scored a grade 5 or higher, find the probability that both students scored a grade 6.

[1 mark]

[1 mark]

[2 marks]

[2 marks]

[3 marks]

Markscheme (M1)(M1)

Note: Award (M1) for seen, (M1) for multiplying their first probability by .

OR

Note: Award (M1) for seen, (M1) for dividing their first probability by .

(A1)(ft) (G3)

Note: Follow through from part (d).

[3 marks]

×10their 26

949

10their 26

949

×1050

949

2650

×1050

949

their 2650

= (0.0706, 0.0706436… , 7.06436… %)45637

4g.

The number of minutes that the 50 students spent preparing for the test was normally distributed with a mean of 105 minutes and astandard deviation of 20 minutes.

Calculate the probability that a student chosen at random spent at least 90 minutes preparing for the test.

Markscheme (M1)

OR

(M1)

Note: Award (M1) for a diagram showing the correct shaded region .

(A1) (G2)

[2 marks]

P(X ⩾ 90)

(> 0.5)

0.773 (0.773372…) 0.773 (0.773372… , 77.3372… %)

4h. Calculate the expected number of students that spent at least 90 minutes preparing for the test.

[2 marks]

[2 marks]

Markscheme (M1)

(A1)(ft) (G2)

Note: Follow through from part (f)(i).

[2 marks]

0.773372… × 50

= 38.7 (38.6686…)

5a.

In a school, all Mathematical Studies SL students were given a test. The test contained four questions, each one on a different topicfrom the syllabus. The quality of each response was classified as satisfactory or not satisfactory. Each student answered only three ofthe four questions, each on a separate answer sheet.

The table below shows the number of satisfactory and not satisfactory responses for each question.

If the teacher chooses a response at random, find the probability that it is a response to the Calculus question;

Markscheme (A1)(A1)(G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

( ; 0.2; 20%)15

1890

5b. If the teacher chooses a response at random, find the probability that it is a satisfactory response to the Calculus question;

Markscheme (A1)(A1)(G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

( ; 0.1̄; 0.111111… ; 11.1%)19

1090

5c. If the teacher chooses a response at random, find the probability that it is a satisfactory response, given that it is a response to theCalculus question.

[2 marks]

[2 marks]

[2 marks]

Markscheme (A1)(A1)(G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

( ; 0.5̄; 0.555556… ; 55.6%)59

1018

5d. The teacher groups the responses by topic, and chooses two responses to the Logic question. Find the probability that both are notsatisfactory.

Markscheme (A1)(M1)

Note: Award (A1) for two correct fractions seen, (M1) for multiplying their two fractions.

(A1)(G2)

[3 marks]

×620

519

( ; 0.0789473… ; 7.89%)338

30380

5e.

A test is carried out at the 5% significance level for the data in the table.

State the null hypothesis for this test.

Markscheme: quality (of response) and topic (from the syllabus) are independent (A1)

Note: Accept there is no association between quality (of response) and topic (from the syllabus). Do not accept “not related” or “notcorrelated” or “influenced”.

[1 mark]

χ2

H0

5f. Show that the expected frequency of satisfactory Calculus responses is 12.

MarkschemeOR (M1)

Note: Award (M1) for correct substitution in expected value formula.

(AG)

Note: The conclusion, , must be seen for the (A1) to be awarded.

[1 mark]

× × 901890

6090

18×6090

(=) 12

(=) 12

[3 marks]

[1 mark]

[1 mark]

5g. Write down the number of degrees of freedom for this test.

Markscheme3 (A1)

[1 mark]

5h. Use your graphic display calculator to find the statistic for this data.

Markscheme (G2)

[2 marks]

χ2

(χ2calc =) 1.46 (1.46¯̄¯̄¯̄36 ; 1.46363…)

5i.

The critical value for this test is 7.815.

State the conclusion of this test. Give a reason for your answer.

MarkschemeOR (R1)

the null hypothesis is not rejected (A1)(ft)

OR

the quality of the response and the topic are independent (A1)(ft)

Note: Award (R1) for a correct comparison of either their statistic to the critical value or the correct -value 0.690688… to thetest level, award (A1)(ft) for the correct result from that comparison. Accept “ ” for the comparison, but only if their valueis explicitly seen in part (f). Follow through from their answers to part (f) and part (c). Do not award (R0)(A1).

[2 marks]

χ2

1.46 < 7.815 0.690688… > 0.05

χ2 χ2 p

χ2calc < χ2

crit χ2calc

6a.

Consider these three propositions, in which x is a natural number.

Write down in symbolic form the compound proposition

“If is a factor of 60 then is a multiple of 5 or is not a multiple of 4.”

p: x is a factor of 60q: x is a multiple of 4r: x is a multiple of 5

x x x

[1 mark]

[2 marks]

[2 marks]

[3 marks]

Markscheme (A1)(A1)(A1)

Note: Award (A1) for “ ”.

Award (A1) for “ ” or “ ” (or “ ”or “ ”)

Award (A1) for “ ”.

Award at most (A1)(A1)(A0) if parentheses are missing for .

Award (A0)(A0)(A1) for .

[3 marks]

p ⇒ (r ∨ ¬q)

p ⇒

r ∨ ¬q r ∨ q ¬q ∨ r q ∨ r

¬q

r ∨ ¬q

(p ⇒ r) ∨ ¬q

6b. Write down in words the compound proposition .

Markscheme is not a multiple of 5 and ( ) is (either) a factor of 60 or ( ) is a multiple of 4, but not both (A1)(A1)(A1)

Note: Award (A1) for “ is not a multiple of 5”, (A1) for “( ) is a factor of 60 or ( ) is a multiple of 4 but not both”, (A1) for “and” in thecorrect position. Accept only “but not both” in the second (A1).

Award at most (A1)(A1)(A0) for using extra statements such as “If ...then”, “if and only if” etc.

[3 marks]

¬r ∧ (p∨

q)−

x x x

x x x

6c. Copy the following truth table and complete the last three columns.

[3 marks]

[3 marks]

Markscheme

(A1)(A1)(A1)(ft)

Note: Award (A1) for each correct column. Last column follows through from previous two.

[3 marks]

6d. State why the compound proposition is not a logical contradiction.

Markschemebecause not all the entries in the column are F (R1)(ft)

Note: If all entries in the last column of their truth table are T, award (R1)(ft) for an answer of “it is a tautology”. Only award (R1)(ft) ifthe column is identified in the justification.

[1 mark]

¬r ∧ (p∨

q)−

¬r ∧ (p∨

q)−

6e. A row from the truth table from part (c) is given below.

Write down one value of that satisfies these truth values.

Markschemeaccept one of: 1 OR 2 OR 3 OR 6 (A1)

Note: Award (A1) for any one of the above answers.

[1 mark]

x

[1 mark]

[1 mark]

7a.

The manager of a folder factory recorded the number of folders produced by the factory (in thousands) and the production costs (inthousand Euros), for six consecutive months.

Draw a scatter diagram for this data. Use a scale of 2 cm for 5000 folders on the horizontal axis and 2 cm for 10 000 Euros on thevertical axis.

Markscheme

(A4)

Notes: Award (A1) for correct scales and labels. Award (A0) if axes are reversed and follow through for their points.

Award (A3) for all six points correctly plotted, (A2) for four or five points correctly plotted, (A1) for two or three points correctly plotted.

If graph paper has not been used, award at most (A1)(A0)(A0)(A0). If accuracy cannot be determined award (A0)(A0)(A0)(A0).

[4 marks]

7b. Write down, for this set of data the mean number of folders produced, ;

Markscheme (A1)(G1)

[1 mark]

x̄

(x̄ =) 21

7c. Write down, for this set of data the mean production cost, .C̄

[4 marks]

[1 mark]

[1 mark]

Markscheme (A1)(G1)

Note: Accept (i) 21000 and (ii) 55000 seen.

[1 mark]

(C̄ =) 55

7d. Label the point on the scatter diagram.

Markschemetheir mean point M labelled on diagram (A1)(ft)(G1)

Note: Follow through from part (b).

Award (A1)(ft) if their part (b) is correct and their attempt at plotting in part (a) is labelled M.

If graph paper not used, award (A1) if is labelled. If their answer from part (b) is incorrect and accuracy cannot be determined,award (A0).

[1 mark]

M(x̄, C̄)

(21, 55)

(21, 55)

7e. Use your graphic display calculator to find the Pearson’s product–moment correlation coefficient, .

Markscheme (G2)

Note: Award (G2) for 0.99 seen. Award (G1) for 0.98 or 0.989. Do not accept 1.00.

[2 marks]

r

(r =) 0.990 (0.989568…)

7f. State a reason why the regression line on is appropriate to model the relationship between these variables.

Markschemethe correlation coefficient/r is (very) close to 1 (R1)(ft)

OR

the correlation is (very) strong (R1)(ft)

Note: Follow through from their answer to part (d).

OR

the position of the data points on the scatter graphs suggests that the tendency is linear (R1)(ft)

Note: Follow through from their scatter graph in part (a).

[1 mark]

C x

[1 mark]

[2 marks]

[1 mark]

7g. Use your graphic display calculator to find the equation of the regression line on .

Markscheme (G2)

Notes: Award (G1) for , (G1) for 14.2.

Award a maximum of (G0)(G1) if the answer is not an equation.

Award (G0)(G1)(ft) if gradient and -intercept are swapped in the equation.

[2 marks]

C x

C = 1.94x + 14.2 (C = 1.94097… x + 14.2395…)

1.94x

C

7h. Draw the regression line on on the scatter diagram.

Markschemestraight line through their (A1)(ft)

-intercept of the line (or extension of line) passing through (A1)(ft)

Notes: Follow through from part (f). In the event that the regression line is not straight (ruler not used), award (A0)(A1)(ft) if linepasses through both their and , otherwise award (A0)(A0). The line must pass through the midpoint, not near thispoint. If it is not clear award (A0).

If graph paper is not used, award at most (A1)(ft)(A0).

[2 marks]

C x

M(21, 55)

C 14.2 (±1)

(21, 55) (0, 14.2)

7i.

Every month the factory sells all the folders produced. Each folder is sold for 2.99 Euros.

Use the equation of the regression line to estimate the least number of folders that the factory needs to sell in a month to exceedits production cost for that month.

[2 marks]

[2 marks]

[4 marks]

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© International Baccalaureate Organization 2018

International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markscheme (M1)(M1)

Note: Award (M1) for seen and (M1) for equating to their equation of the regression line. Accept an inequality sign.

Accept a correct graphical method involving their part (f) and .

Accept drawn on their scatter graph.

(this step may be implied by their final answer) (A1)(ft)(G2)

(A1)(ft)(G3)

Note: Follow through from their answer to (f). Use of 3 sf gives an answer of .

Award (G2) for or 13.524 or a value which rounds to 13500 seen without workings.

Award the last (A1)(ft) for correct multiplication by 1000 and an answer satisfying revenue > their production cost.

Accept 13.6 thousand (folders).

[4 marks]

2.99x = 1.94097… x + 14.2395…

2.99x

2.99x

C = 2.99x

x = 13.5739…

13 600 (13 574)

13 524

13.5739…