New test - January 28, 2016 · were some who gave the frequency values on the y-axis, rather than the wages on the x-axis, as their quartiles and interquartile range. a b c 330 400

New test - January 28, 2016 [133 marks]

1a. [4 marks]

The weekly wages (in dollars) of 80 employees are displayed in the cumulative frequency curve below.

(i) Write down the median weekly wage.

(ii) Find the interquartile range of the weekly wages.

Markscheme(i) median weekly wage (dollars) A1 N1

(ii) lower quartile , upper quartile (A1)(A1)

(dollars) (accept any notation suggesting interval to ) A1 N3

Note: Exception to the FT rule. Award A1(FT) for an incorrect IQR only if both quartiles are explicitly noted.

[4 marks]

Examiners reportMany candidates answered this question completely correctly, earning full marks in all parts of the question. In parts (a) and (b), therewere some who gave the frequency values on the y-axis, rather than the wages on the x-axis, as their quartiles and interquartile range.

= 400

= 330 = 470

IQR = 140 330 470

1b. [3 marks]The box-and-whisker plot below displays the weekly wages of the employees.

Write down the value of

(i) ;

(ii) ;

(iii) .

Markscheme(i) (dollars) A1 N1

(ii) (dollars) A1 N1

(iii) (dollars) A1 N1

[3 marks]

Examiners reportMany candidates answered this question completely correctly, earning full marks in all parts of the question. In parts (a) and (b), therewere some who gave the frequency values on the y-axis, rather than the wages on the x-axis, as their quartiles and interquartile range.

a

b

c

330

400

700

1c. [3 marks]Employees are paid per hour.

Find the median number of hours worked per week.

Markschemevalid approach (M1)

e.g.

correct substitution (A1)

e.g.

median hours per week A1 N2

[3 marks]

Examiners reportFor part (c), the majority of candidates seemed to understand what was required, though there were a few who used an extreme valuesuch as , rather than the median value.

$ 20

hours =wagesrate

40020

= 20

700

1d. [5 marks]Employees are paid per hour.

Find the number of employees who work more than hours per week.

$20

25

Markschemeattempt to find wages for 25 hours per week (M1)

e.g.

correct substitution (A1)

e.g.

finding wages (A1)

65 people (earn 500 ) (A1)

15 people (work more than 25 hours) A1 N3

[5 marks]

Examiners reportIn part (d), some candidates simply answered , which is the number of workers earning or less, rather than finding the numberof workers who earned more than . It was interesting to note that quite a few candidates gave their final answer as , rather than

.

wages = hours × rate

25 × 20

= 500

⩽

65 $500$500 14

15

[2 marks]2a.

The cumulative frequency curve below represents the marks obtained by 100 students.

Find the median mark.

Markschemeevidence of median position (M1)

e.g. 50, line on sketch

median is 56 A1 N2

[2 marks]

Examiners reportOverall, this question was done well by candidates. In part (a), a surprising number of candidates found the median position (thecumulative frequency) on the y-axis, but did not find the median mark on the x-axis.

[3 marks]2b. Find the interquartile range.

Markschemelower quartile , upper quartile (A1)(A1)

interquartile range A1 N3

[3 marks]

Examiners reportOverall, this question was done well by candidates. In part (a), a surprising number of candidates found the median position (thecumulative frequency) on the y-axis, but did not find the median mark on the x-axis. Similar misunderstanding was shown by somecandidates in part (b), when attempting to find the interquartile range.

= 40 = 70

= 30

[2 marks]3a.

The ages of people attending a music concert are given in the table below.

Find p .

Markschemeevidence of valid approach (M1)

e.g. , line on graph at

A1 N2

[2 marks]

Examiners reportPart (a) was generally answered correctly, with most candidates showing a good grasp of cumulative frequency from a table.

92 + 52 x = 31

p = 144

3b. [5 marks]The cumulative frequency diagram is given below.

Use the diagram to estimate

(i) the 80th percentile;

(ii) the interquartile range.

Markscheme(i) evidence of valid approach (M1)

e.g. line on graph, , using complement

A1 N2

(ii) ; (A1)(A1)

(accept any notation that suggests an interval) A1 N3

[5 marks]

Examiners reportA surprising number of candidates had difficulty reading values off the cumulative frequency curve. A common incorrect answer for(b)(i) was 29, indicating carelessness with the given scale. Too many candidates gave 40 and 120 for the quartile values.

0.8 × 160

= 29.5

= 23Q1 = 29Q3

IQR = 6

4a. [3 marks]

The histogram below shows the time T seconds taken by 93 children to solve a puzzle.

The following is the frequency distribution for T .

(i) Write down the value of p and of q .

(ii) Write down the median class.

Markscheme(i) , A1A1 N2

(ii) A1 N1

[3 marks]

Examiners reportParts (a) and (b) were generally well done. The terms "median" and "median class" were often confused.

p = 17 q = 11

75 ≤ T < 85

[2 marks]4b. A child is selected at random. Find the probability that the child takes less than 95 seconds to solve the puzzle.

Markschemeevidence of valid approach (M1)

e.g. adding frequencies

A1 N2

[2 marks]

= 0.8172043…7693

P(T < 95) = = 0.8177693

Examiners reportParts (a) and (b) were generally well done. The terms "median" and "median class" were often confused.

4c. [2 marks]Consider the class interval .

(i) Write down the interval width.

(ii) Write down the mid-interval value.

Markscheme(i) 10 A1 N1

(ii) 50 A1 N1

[2 marks]

Examiners reportIn part (c) some candidates had problems with the term "interval width" and there were some rather interesting mid-interval valuesnoted.

45 ≤ T < 55

4d. [4 marks]Hence find an estimate for the

(i) mean;

(ii) standard deviation.

Markscheme(i) evidence of approach using mid-interval values (may be seen in part (ii)) (M1)

A2 N3

(ii)

A1 N1

[4 marks]

Examiners reportIn part (d), candidates often ignored the "hence" command and estimated values from the graph rather than from the information inpart (c).

79.1397849

= 79.1x̄̄̄

16.4386061

σ = 16.4

4e. [2 marks]John assumes that T is normally distributed and uses this to estimate the probability that a child takes less than 95 seconds tosolve the puzzle.

Find John’s estimate.

Markschemee.g. standardizing,

A1 N2

[2 marks]

z = 0.9648…

0.8326812

P(T < 95) = 0.833

Examiners reportThose who correctly obtained the mean and standard deviation had little difficulty with part (e) although candidates often usedunfamiliar calculator notation as their working or used the mid-interval value as the mean of the distribution.

[1 mark]5a.

The following is a cumulative frequency diagram for the time t, in minutes, taken by 80 students to complete a task.

Write down the median.

Markschememedian A1 N1

[1 mark]

Examiners reportThis question was answered successfully by a majority of candidates. A common error was to use values of 20 and 60 for the lowerand upper quartiles. Some were careless when reading the graph scale and wrote incorrect answers as a result.

m = 32

[3 marks]5b. Find the interquartile range.

Markschemelower quartile , upper quartile (A1)(A1)

A1 N3

[3 marks]

= 22Q1 = 40Q3

interquartile range = 18


5c. [2 marks]Complete the frequency table below.

Markscheme

A1A1 N2

[2 marks]


6a. [3 marks]

In a school with 125 girls, each student is tested to see how many sit-up exercises (sit-ups) she can do in one minute. The results aregiven in the table below.

(i) Write down the value of p.

(ii) Find the value of q.

Markscheme(i) A1 N1

(ii) for evidence of using sum is 125 (or ) (M1)

A1 N2

[3 marks]

Examiners reportPart (a) of this question was well done.

p = 65

99 − p

q = 34

[2 marks]6b. Find the median number of sit-ups.

Markschemeevidence of median position (M1)

e.g. 63rd student,

median is 17 (sit-ups) A1 N2

[2 marks]

Examiners reportFinding the median seemed to be the most difficult for the candidates. Most had the idea that it was in the middle but did not knowhow to find the value.

1252

[2 marks]6c. Find the mean number of sit-ups.

Markschemeevidence of substituting into (M1)

e.g. ,

mean A1 N2

[2 marks]

Examiners reportWhen calculating the mean, many ignored the frequencies.

∑ fx

125

15(11)+16(21)+17(33)+18(34)+19(18)+20(8)125

2176125

= 17.4

[2 marks]7a.

A box contains 100 cards. Each card has a number between one and six written on it. The following table shows the frequencies foreach number.

Calculate the value of k.

Markschemeevidence of using (M1)

A1 N2

[2 marks]

Examiners reportFrequencies and median seemed well understood, but quartiles and inter-quartile range less so.

∑ = 100fi

k = 4

7b. [5 marks]Find

(i) the median;

(ii) the interquartile range.

Markscheme(i) evidence of median position (M1)

e.g. 50th item,

A1 N2

(ii) and (A1)(A1)

(accept 1 to 5 or , etc.) A1 N3

[5 marks]

Examiners reportFrequencies and median seemed well-understood, but quartiles and interquartile range less so. A few students, probably based on pastpapers, drew cumulative frequency diagrams, generating slightly different answers for median and quartiles.

26 + 10 + 20 = 56

median = 3

= 1Q1 = 5Q3

interquartile range = 4 5 − 1

[3 marks]8a.

A fisherman catches 200 fish to sell. He measures the lengths, l cm of these fish, and the results are shown in the frequency tablebelow.

Calculate an estimate for the standard deviation of the lengths of the fish.

Markschemeevidence of using mid-interval values (5, 15, 25, 35, 50, 67.5, 87.5) (M1)

(cm) A2 N3

[3 marks]

Examiners reportPart (a) defeated the vast majority of candidates who clearly had not been taught data entry. Some schools had attempted to teach howto use a formula rather than the GDC to find the standard deviation and their students invariably used this formula incorrectly. Use ofthe GDC was not only expected but should be emphasized as stated in the syllabus.

σ = 19.8

8b. [6 marks]A cumulative frequency diagram is given below for the lengths of the fish.

Use the graph to answer the following.

(i) Estimate the interquartile range.

(ii) Given that of the fish have a length more than , find the value of k.

Markscheme(i) , (A1)(A1)

(accept any notation that suggests the interval 15 to 40) A1 N3

(ii) METHOD 1

have a length less than k (A1)

(A1)

(cm) A1 N2

METHOD 2

(A1)

(A1)

(cm) A1 N2

[6 marks]

Examiners reportPart (b) revealed poor understanding of cumulative frequency and the IQR was often reported as an interval.

40% k cm

= 15Q1 = 40Q3

IQR = 25

60%

0.6 × 200 = 120

k = 30

0.4 × 200 = 80

200 − 80 = 120

k = 30

8c. [2 marks]In order to sell the fish, the fisherman classifies them as small, medium or large.

Small fish have a length less than .

Medium fish have a length greater than or equal to but less than .

Large fish have a length greater than or equal to .

Write down the probability that a fish is small.

Markscheme (M1)

A1 N2

[2 marks]

Examiners reportThis was generally answered well although a number of candidates had difficulty with using the formula for expected value.

20 cm

20 cm 60 cm

60 cm

l < 20 cm ⇒ 70 fish

P(small) = (= 0.35)70200

8d. [2 marks]The cost of a small fish is , a medium fish , and a large fish .

Copy and complete the following table, which gives a probability distribution for the cost .

Markscheme

A1A1 N2

[2 marks]


$4 $10 $12

$X

[2 marks]8e. Find .

Markschemecorrect substitution (of their p values) into formula for (A1)

e.g.

(accept ) A1 N2

[2 marks]


E(X)

E(X)

4 × 0.35 + 10 × 0.565 + 12 × 0.085

E(X) = 8.07 $8.07

[1 mark]9a.

The following diagram is a box and whisker plot for a set of data.

The interquartile range is 20 and the range is 40.

Write down the median value.

Markscheme18 A1 N1

[1 mark]

Examiners reportMost candidates were able to find the values for the median, lower quartile, and point b. A large majority answered this questioncorrectly.

9b. [4 marks]Find the value of

(i) ;

(ii) .

Markscheme(i) 10 A2 N2

(ii) 44 A2 N2

[4 marks]

Examiners reportMost candidates were able to find the values for the median, lower quartile, and point b. A large majority answered this questioncorrectly.

a

b

[2 marks]10a.

A standard die is rolled 36 times. The results are shown in the following table.

Write down the standard deviation.

Markscheme A2 N2

[2 marks]

Examiners reportSurprisingly, this question was not answered well primarily due to incorrect GDC use and a lack of understanding of the terms"median" and "interquartile range". Many candidates opted for an analytical approach in part (a) which always resulted in mistakes.

σ = 1.61

[1 mark]10b. Write down the median score.

Markschememedian A1 N1

[1 mark]

Examiners reportSome candidates wrote the down the mean instead of the median in part (b).

= 4.5

[3 marks]10c. Find the interquartile range.

Markscheme , (may be seen in a box plot) (A1)(A1)

(accept any notation that suggests the interval 3 to 5) A1 N3

[3 marks]

Examiners reportSurprisingly, this question was not answered well primarily due to incorrect GDC use and a lack of understanding of the terms"median" and "interquartile range".

= 3Q1 = 5Q3

IQR = 2

[4 marks]11a.

The following frequency distribution of marks has mean 4.5.

Find the value of x.

Markscheme , (seen anywhere) A1

evidence of substituting into mean (M1)

correct equation A1

e.g. ,

A1 N2

[4 marks]

Examiners reportSurprisingly, this question was not well done by many candidates. A good number of candidates understood the importance of thefrequencies in calculating mean. Some neglected to sum the frequencies for the denominator, which often led to a negative value for afrequency. Unfortunately, candidates did not appreciate the unreasonableness of this result.

∑ fx = 1(2) + 2(4) + … + 7(4) ∑ fx = 146 + 5x

∑ fx

∑ f

= 4.5146+5x

34+x146 + 5x = 4.5(34 + x)

x = 14

[2 marks]11b. Write down the standard deviation.

Markscheme A2 N2

[2 marks]

Examiners reportSurprisingly, this question was not well done by many candidates. A good number of candidates understood the importance of thefrequencies in calculating mean. Some neglected to sum the frequencies for the denominator, which often led to a negative value for afrequency. Unfortunately, candidates did not appreciate the unreasonableness of this result. In part (b), many candidates could notfind the standard deviation in their GDC, often trying to calculate it by hand with no success. Further, many could not distinguishbetween the sample and the population standard deviation given in the GDC.

σ = 1.54

12a. [4 marks]

The following table gives the examination grades for 120 students.

Find the value of

(i) p ;

(ii) q .

Markscheme(a) (i) evidence of appropriate approach (M1)

e.g. ,

A1 N2

(ii) evidence of valid approach (M1)

e.g. their value of p,

A1 N2

[4 marks]

Examiners reportThe majority of candidates had little trouble finding the missing values in the frequency distribution table.

9 + 25 + 35 34 + 35

p = 69

109− 120 − (9 + 25 + 35 + 11)

q = 40

[2 marks]12b. Find the mean grade.

Markschemeevidence of appropriate approach (M1)

e.g. substituting into , division by 120

mean A1 N2

[2 marks]

∑ fx

n

= 3.16

Examiners reportMany did not seem comfortable calculating the mean and standard deviation using their GDCs.

The correct mean was often found without the use of the statistical functions on the graphing calculator, but a large number ofcandidates were unable to find the standard deviation.

[1 mark]12c. Write down the standard deviation.

Markscheme1.09 A1 N1

[1 mark]

Examiners reportMany did not seem comfortable calculating the mean and standard deviation using their GDCs.

The correct mean was often found without the use of the statistical functions on the graphing calculator, but a large number ofcandidates were unable to find the standard deviation.

[2 marks]13a.

There are nine books on a shelf. For each book, x is the number of pages, and y is the selling price in pounds (£). Let r be thecorrelation coefficient.

Write down the possible minimum and maximum values of r .

Markschememin value of r is , max value of r is 1 A1A1 N2

[2 marks]

Examiners report[N/A]

−1

13b. [1 mark]Given that , which of the following diagrams best represents the data.

MarkschemeC A1 N1

[1 mark]


r = 0.95

13c. [2 marks]For the data in diagram D , which two of the following expressions describe the correlation between x and y ?

perfect, zero, linear, strong positive, strong negative, weak positive, weak negative

Markschemelinear, strong negative A1A1 N2

[2 marks]


[1 mark]14a.

The cumulative frequency curve below represents the heights of 200 sixteen-year-old boys.

Use the graph to answer the following.

Write down the median value.

Markscheme A1 N1

[1 mark]

Examiners reportParts (a) and (b) were generally well done.

median = 174(cm)

[2 marks]14b. A boy is chosen at random. Find the probability that he is shorter than .

Markschemeattempt to find number shorter than 161 (M1)

e.g. line on graph, 12 boys

A1 N2

[2 marks]

Examiners reportParts (a) and (b) were generally well done.

161 cm

p = (= 0.06)12200

[3 marks]14c. Given that of the boys are taller than , find h .82% h cm

MarkschemeMETHOD 1

have a height less than h (A1)

(36 may be seen as a line on the graph) (A1)

(cm) A1 N2

METHOD 2

(164 may be seen as a line on the graph) (A1)

(A1)

(cm) A1 N2

[3 marks]

Examiners reportSome candidates could only earn the first mark in part (c) for finding of 200. Others gave the answer as 164, neglecting tosubtract this value from the total of 200.

18%

0.18 × 200 = 36

h = 166

0.82 × 200 = 164

200 − 164 = 36

h = 166

82%

[3 marks]15a.

A scientist has 100 female fish and 100 male fish. She measures their lengths to the nearest cm. These are shown in the following boxand whisker diagrams.

Find the range of the lengths of all 200 fish.

Markschemecorrect end points (A1)(A1)

max = 27 , min = 4

range = 23 A1 N3

[3 marks]

Examiners reportWhile there were a large number of candidates who answered both parts of this question correctly, a surprising number did not knowhow to find the range of all 200 fish in part (a). Common errors included finding the ranges of the male and female fish separately, oraveraging the separate ranges of the male and female fish.

15b. [2 marks]Four cumulative frequency graphs are shown below.

Which graph is the best representation of the lengths of the female fish?

MarkschemeGraph 3 A2 N2

[2 marks]

Examiners reportSome candidates did not interpret the cumulative frequency graphs correctly, or just seemed to guess which graph was correct. Themost common incorrect "guess" was graph 4, likely because this graph had a more familiar cumulative shape.

16a. [2 marks]

A data set has a mean of 20 and a standard deviation of 6.

Each value in the data set has 10 added to it. Write down the value of

(i) the new mean;

(ii) the new standard deviation.

Markscheme(i) new mean is A1 N1

(ii) new sd is 6 A1 N1

[2 marks]


20 + 10 = 30

16b. [3 marks]Each value in the original data set is multiplied by 10.

(i) Write down the value of the new mean.

(ii) Find the value of the new variance.

Markscheme(i) new mean is A1 N1

(ii) METHOD 1

variance is 36 A1

new variance is A1 N2

METHOD 2

new sd is 60 A1

new variance is A1 N2

[3 marks]


20 × 10 = 200

36 × 100 = 3600

= 3600602

[2 marks]17a.

Each day, a factory recorded the number ( ) of boxes it produces and the total production cost ( ) dollars. The results for nine daysare shown in the following table.

Write down the equation of the regression line of y on x .

Markscheme A1A1 N2

[2 marks]


x y

y = 10.7x + 121

17b. [2 marks]Use your regression line from part (a) as a model to answer the following.

Interpret the meaning of

(i) the gradient;

(ii) the y-intercept.

Markscheme(i) additional cost per box (unit cost) A1 N1

(ii) fixed costs A1 N1

[2 marks]


[2 marks]17c. Estimate the cost of producing 60 boxes.

Printed for North Hills Preparatory

© International Baccalaureate Organization 2016 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markschemeattempt to substitute into regression equation M1

e.g. ,

(accept from 3 s.f. values) A1 N2

[2 marks]


y = 10.7 × 60 + 121 y = 760.12…

cost = $760 $763

17d. [3 marks]The factory sells the boxes for $19.99 each. Find the least number of boxes that the factory should produce in one day in orderto make a profit.

Markschemesetting up inequality (accept equation) M1

e.g.

A1

13 boxes (accept 14 from , using 3 s.f. values) A1 N2

Note: Exception to the FT rule: if working shown, award the final A1 for a correct integer solution for their value of x.

[3 marks]


19.99x > 10.7x + 121

x > 12.94…

x > 13.02

17e. [4 marks]Comment on the appropriateness of using your model to

(i) estimate the cost of producing 5000 boxes;

(ii) estimate the number of boxes produced when the total production cost is $540.

Markscheme(i) this would be extrapolation, not appropriate R1R1 N2

(ii) this regression line cannot predict x from y, not appropriate R1R1 N2

[4 marks]


Documents

New test - January 28, 2016 · were some who gave the frequency values on the y-axis, rather than the wages on the x-axis, as their quartiles and interquartile range. a b c 330 400