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binomcdf(22,0.1,1,4)

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23

Copyright © Insight Publications 2017 SECTION B

Question 2a.i.

Worked solution

20

2

20

3

2

0

3

2

( ) 120

2( ) .

2

1 2

3

1 2 2.

3 3

2120, 180

3

a

a

a

E X

xE X x dx

a

xdx

a

x

a

a a

a

aa

Mark allocation: 2 marks

1 method mark for setting up the integral for E(X)

1 answer mark for the correct antiderivative and evaluation leading to 180a

Tip

This is a ‘show that’ question, so appropriate working must be shown.

24


Question 2a.ii.

Worked solution

180

2150

2Pr( 150)

180

xX dx

Using CAS gives 11

Pr( 150) .36

X


1 method mark for setting up the integral

1 answer mark for 11

36

Tip

This question is worth 2 marks, so you must also provide a solution step.

25


Question 2b.

Worked solution

RF ~ ( 25, 5)X N

Using CAS gives RFPr( 18) 0.0808.X


1 method mark for RF ~ ( 25, 5)X N

1 answer mark for 0.0808

Tip

Be careful to define the distribution using a new variable. Something like

RFX is suitable.

26


Question 2c.

Worked solution

Using symmetry, the mean is 24 mm.

JJ ~ ( 24, ?)X N , where JJPr( 20) 0.1.X

Using CAS gives 3.12.


1 answer mark for finding the mean

1 method mark for setting up an integral to find the standard deviation

1 answer mark for the correct standard deviation

Tip

Be careful to use a different notation for defining the distribution.

27


Question 2d.i.

Worked solution

JJ

JJ RF

Pr( 18)Pr(JJ | length 18)

Pr(JJ) Pr( 18) Pr(RF) Pr( 18)

0.7 0.027235

0.7 0.027235 0.3 0.0808

0.440

X

X X


1 method mark for recognising conditional probability or for using a tree diagram and

for finding JJPr( 18) 0.7 0.027235X

1 method mark for having a denominator that involves 0.7 multiplied by one

probability and 0.3 multiplied by another

1 answer mark for 0.440

28


Question 2d.ii.

Worked solution

0.7, 500

(1 ) 0.7 0.3ˆ( ) 0.7,

500

ˆ ~ ( 0.7, 0.0205)

p n

p pE p p

n

P N

ˆPr( 0.75) 0.0073P


1 mark for identifying parameters 0.7, 0.0205

1 mark for answer ˆPr( 0.75) 0.0073P (although if the rounded value for is used

this will give ˆPr( 0.75) 0.0074,P so accept either 0.0073 or 0.0074)

VCE Mathematical Methods Units 3&4 Trial Examination 2 Suggested Solutions

14 MMU34EX2_SS_2017.FM Copyright © 2017 Neap

d. i. Graphically, a translation of 3 units right, or 9 units to the left to match the

graphs is required.

As d > 0, translation must be to the left, therefore d = 9 is the smallest

possible value. A1

ii. Starting with a translation of 9 units left and then multiples of the period

(12 months) gives the following general solution for d.

d = 9 + 12k where k ∈ Z+ ∪ {0} expression for d A1

positive integers for k A1

e.

M1

Question 3 (19 marks)

a. i.

A1

w t( ) 1000 200π t

6-----

q t 12–( )+sin+=

w′ t( )100π

3------------

π t

6-----

q+cos=

w′ t( ) 0>

100π3

------------π t

6-----

q+cos 0>

A1

minimum value for πt

6----- cos 1–=

100π3

------------– q 0>+

q100π

3------------>

A1

X Bi n p,( )∼

X Bi 201

10------,

∼ M1

Pr X 3<( ) Pr X 2≤( )=

0.6769=


Copyright © 2017 Neap MMU34EX2_SS_2017.FM 15

ii.

b.

c. i.

ii.

iii.

To halve the standard deviation, increase n by a factor of 4.

A1

X Bi 201

10------,

∼

Pr X = 0 X 2≤( )Pr X = 0 X 2≤∩( )

Pr X 2≤( )--------------------------------------------=

Pr X = 0( )

Pr X 2≤( )-----------------------=

0.121576…

0.676926…----------------------------=

0.1796=

M1

A1

X Bi n1

10------,

∼ n ?=

Pr X 0>( ) 0.8≥

1 Pr X = 0( ) 0.8≥–

1 0.9n

0.8≥–

n 15.27≥

n 16=

M1

A1

E P̂( ) p=

1

10------= A1

sd P̂( )p 1 p–( )

n--------------------=

1

10------ 1

1

10------–

20---------------------------=

0.0671 (to four decimal places)≈ A1

sd P̂( ) ∝ 1

n-------

n 4 20×=

80=


16 MMU34EX2_SS_2017.FM Copyright © 2017 Neap

d. i.

M1

ii.

Note: Students must use this step as it is a ‘hence’ question.

e.

An approximate 95% confidence interval for p is given by:

M1

a = 3, b = 7 A1

Y N µ σ2

,( )∼

Y N 2 0.42

,( )∼ M1

Pr 1.5 Y 2< <( ) 0.3944 (to four decimal places)≈

Pr Y 2 Y 1.5≥≥( )Pr Y 2>( )

Pr Y 1.5>( )---------------------------=

Pr Y 2>( )

Pr 1.5 Y 2< <( ) Pr Y 2>( )+------------------------------------------------------------------=

A1

M1

0.5

0.3944 0.5+------------------------------=

0.5591= A1

p̂20

100---------=

0.2=

p̂ zp̂ 1 p̂–( )

n-------------------- p p̂ z

p̂ 1 p̂–( )

n--------------------+< <–

Where p̂ 0.2 z, 2 n, 100.= = =

0.2 20.2 0.8×

100--------------------- 0.2 2

0.2 0.8×

100---------------------+,– 3

25------

7

25------,

=


Copyright © 2017 Neap MMU34EX2_SS_2017.FM 17

f.

correct simultaneous equations M1

solving simultaneously gives:

µ = 2.1232 A1

σ = 0.4863 A1


a. dilation factor of from the x-axis or alternatively a dilation factor of 2 from

the y-axis both correct A1

translation of 5 in the negative y-direction A1

b. finding correct x-coordinates M1

Pr X 2>( )12

20------=

0.6=

Pr 1.5 X 2< <( )6

20------=

0.3=

x

0.1

0.3

0.6

1.5 2

zz2

z1

zx µ–

σ------------=

0.25334–2 µ–

σ------------=

1.28155–1.5 µ–

σ----------------=

1

4---

1

4---x

25– a

2=

1

4---x

2a

25+=

x2

4 a2

5+( )=

x 2 a2

5+ as x 0>=

B is 2 a2

5+ a a2

,–( ).

Mathematics Methods Trial Exam 2 2016 Solutions Section B Page 14

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Question 3

a.i. 2 245,000, 8,000d

A N

Pr 40,000 0.7340A A1

ii. 5, 0.7340d

Y Bi n p

Pr 3 0.8789Y A1

b. 2?, ?d

B N

Pr 61,000 0.11B 61,000

Pr 61,000 0.89 1.2265B

M1

Pr 42,500 0.20B 42,500

0.8416

M1

solving 50,028 , 8,945

the mean is 50,000 km, the standard deviation is 9,000 km. A1

c. 26

ˆ 0.72 , 36 , 1.9636

p n z

ˆ ˆ ˆ ˆ1 1

ˆ ˆ,p p p p

p z p zn n

0.722 1 0.722 0.722 1 0.722

0.722 1.96 , 0.722 1.9636 36

0.576,0.869 A1

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d.i ˆ 0.7E P p A1

1 0.7 0.3 7ˆvar

36 1200

p pP

n

A1

ii. 36 0.7 25.2np

36 0.7 0.3 2.75npq

2 25.2 2 2.75 19.7,30.7 M1

36, 0.7d

Y Bi n p

Pr 20 30 0.956Y A1





e.i. Since the graph is continuous at 50t

250 50 100 50 50f a b b a A1

Since the total area under the curve is equal to one.

50 100

2

0 50

50 100

3 2

0 50

3 2 2

100 1

1 150 100 1

3 2

1 1 150 50 100 100 100 100 50 50 1

3 2 2

312,500 3 31 , so and

3 312,500 6,250

a t dt b t dt

a t a t t

a

aa b

M1

ii. 40

2

0

3 497Pr 40 1 Pr 40 1

312,500 625T T t dt

expect 497

36 28.6625

accept 28 or 29 A1





iii. graph, correct scale, shape, continuous at 3

50, 50,0.024125

, G1

must show zero for 100t and 0t A1





iv. 50 100

3

0 50

3 3100 55

312,500 6,250E T t dt t t dt A1

50 100

2 4 2

0 50

3 3100 3350

312,500 6,250E T t dt t t dt

22 2var 3350 55 325T E T E T A1

v. sd 325 5 13T

2 55 10 13 18.944,91.056

Pr 2 2 Pr 18.944 91.056T T M1

50 91.056

2

18.94 50

3 3100

312,500 6,250t dt t dt

0.959 A1

vi. Since 50

2

0

3 20.4

312,500 5t dt M1

the median satisfies 50

3100 0.1

6,250

m

t dt

solving gives the median as 54.356 thousand km A1



18


Question 3a.

Worked solution

~ ( 15, 4)

Pr( ) 0.15c

c

X N

X c

Use CAS to find c gives c = 19.145

So c = 191 mm

Mark allocation: 1 mark 1 answer mark for 191 mm

Tip

Take note of the units required. Here the question is given in cm and the answer is required in mm.

Question 3b.

Worked solution

Pr( 9) 0.0668

0.0668 4000 267cX

Mark allocation: 2 marks 1 answer mark for 0.0668 1 answer mark for 267 plants.

Tip

Remember to give your answer as the number of plants, not just the probability.

19


Question 3c.

Worked solution 16

6

( 6)sin 1

10

Using CAS to evaluate the definite integral gives

201

20

xk dx

k

k


1 answer mark for setting 16

6

( 6)sin 1

10

xk dx


6

( 6) 20sin

10

xdx

leading to correct value for k.

Tip

The area under a probability density function is always 1.

Question 3d.

Worked solution

Using the symmetry of the sine curve gives mean as half way between 6 and 16 as 11.

Mark allocation: 1 mark 1 answer mark for the answer of 11.

Tip

When dealing with a symmetrical distribution such as sine, make good use of the symmetry for determining the mean.

20


Question 3e.

Worked solution

6

Pr( ) 0.1

( 60.1 sin

20 10

r

c

X c

xdx

Use CAS to solve gives

8.04

80 mm

c

c

Mark allocation: 2 marks 1 method mark for writing the integral from 6 to c equal to 0.1 1 answer mark for 80 mm

Question 3f.

Worked solution

There are six ways that she could choose exactly two tall plants from the next four.

STTS, SSTT, STST, TSST, TTSS, TSTS

3 1 1 3 1 3 2 1

Pr( ) Pr(TSST)4 3 4 4 4 4 3 33 2 1 1 1 1 3 2

Pr( ) Pr(TTSS)4 3 3 4 4 4 4 33 1 3 1 1 3 1 3

Pr( ) Pr(TSTS)4 3 4 3 4 4 3 4

STTS

SSTT

STST

The sum of these probabilities is 13

48

Mark allocation: 2 marks 1 method mark for determining the six ways: STTS, SSTT, STST, TSST, TTSS and

TSTS


48 must be an exact value

Tip

There are six ways of choosing two tall plants from four because 42 6.C

21


Question 3g.i.

Worked solution

If the sample proportion is ˆ 0.3p and the sample size is 20, then the number of diseased plants in the sample is 0.3 20 =6.

Thus

6 14

ˆPr( 0.3) Pr( 6)

20(0.3) (0.7)

6

0.1916

P X

Mark allocation: 2 marks 1 method mark for finding 6X 1 answer mark for 0.1916

Question 3g.ii.

Worked solution

(1 )ˆ( )

0.3 0.70.1025

20

p psd P

n

Since

0.3 2 0.1025 0.095

0.3 2 0.1025 0.505

we find

ˆPr(0.095 0.505)

Pr(1.9 10.1)

Pr(2 10)

0.9752 using CAS

P

X

X

and ~ ( 20, 0.3)X Bi n p

Mark allocation: 3 marks 1 method mark for finding standard deviation of 0.1025 1 method mark for finding the endpoints of 0.095 and 0.505 1 answer mark for 0.9752

22


Question 3h.

Worked solution

This can be done using CAS.

The 95% confidence interval is (0.428, 0.516).

Mark allocation: 1 mark 1 answer mark for (0.428, 0.516)

Question 3i.

Worked solution

This experiment is a binomial distribution with 26

~ ( 5, ).27

Y Bi n p

Using CAS Pr( 3) 0.0122.X

Mark allocation: 2 marks 1 method mark for recognising the binomial distribution 1 answer mark for 0.0122

10

©THE HEFFERNAN GROUP 2016 Maths Methods 3 & 4 Trial Exam 2 solutions


a. X follows a binomial distribution with n =15 and p =2

9.

Using )5,0,9

2 5,binomCdf(1 ,

...90581.0)5Pr( X 906.0 (correct to 3 decimal places).

(1 mark)

b. i. )ˆ(ˆ of valueexpected PEP

9

2

)proportion population (the

p

(1 mark)

ii. standard deviation of P̂n

pp )1(

135

210

15

9

7

9

2

(1 mark)

c. Solve

2

9´

7

9n

< 0.1 for n. (1 mark)

n > 17.283….

Since n must be an integer (i.e. we are dealing in whole numbers) the minimum

number of customers needed in each sample is 18.

(1 mark)

d. The population proportion is 2

9.

places) decimal 3 o(correct t 652.0

))4,2,9

25,binomCdf(1 (using...65209.0

luesinteger va must take since)42Pr(

...)94348.4...72318.1Pr(

)15...32956.015...11487.0Pr(

...)32956.0ˆ...11487.0Pr(

135

210

9

2ˆ135

210

9

2Pr want We

XX

X

X

P

P

(1 mark)

(1

ma

rk)

(1

mar

k)

(1

mar

k) (1

mar

k)

11

©THE HEFFERNAN GROUP 2016 Maths Methods 3 & 4 Trial Exam 2 solutions

e. i. 15

2

)()(E dttftT

= 9.00498...

= 9.005 minutes (correct to 3 decimal places)

(1 mark)

ii.

15

10

)()10Pr( dttfT


...421391.0

(1 mark)

iii. )10Pr(

)1012Pr()10|12Pr(

T

TTTT (conditional probability formula)


..0.5694545.=

ii)part (from

...421391.0

...239963.0

)10Pr(

)(

)10Pr(

)12Pr(

15

12

T

dttf

T

T

(1 mark)

f. Method 1 – using CAS

Using 1–Prop z interval, with 90.0Level and 2000,8601 Cnx , the 90%

confidence interval for p is (0.920615…, 0.939384…) or (0.921, 0.939) where values

are given correct to 3 decimal places.

(1 mark)

Method 2 – by hand (and a little CAS)

An approximate 90% confidence interval for p is given by

n

ppp

n

ppp

)ˆ1(ˆ64.1ˆ,

)ˆ1(ˆ64.1ˆ

93.0

2000 and )proportion sample (the2000

1860ˆ where

np

The approximate 90% confidence interval for p is therefore

places) decimal 3 correct to (values )939.0,921.0(

...)939356.0...,920643.0(

2000

07.093.064.193.0,

2000

07.093.064.193.0

(1 mark)

(1

mar

k)

(1

mar

k)

(1

mar

k)

EXAMINATION 5 MATHEMATICS METHODS MARKING KEY CALCULATOR-ASSUMED Question 12 (12 marks) Rebecca sells potatoes at her organic fruit and vegetable shop that have weights normally distributed with a mean of 230 g and a standard deviation of 5 g. (a) Determine the probability that one of Rebecca’s potatoes, selected at random, will

weigh between 223 g and 235 g. (1 mark)

Solution 2(230,5 )

(223 235) 0.7606X N

P x

Specific behaviours calculates probability correctly

(b) Five percent of Rebecca’s potatoes weigh less than w g. Determine w to the nearest

gram. (2 marks)

Solution

( ) 0.05221.8 222g to the nearest gram

P X w

w

Specific behaviours gives correct probability statement calculates w correctly

(c) A customer buys twelve potatoes.

(i) Determine the probability that all twelve potatoes weigh between 223 g and 235 g. (2 marks)

Solution

(12,0.7606)( 12) 0.0375

Y bin

P y

Specific behaviours states binomial distribution and its parameters calculates probability correctly

(ii) If the customer is selecting the twelve potatoes one at a time, determine the

probability that it takes the selection of eight potatoes before six potatoes weighing between 223 g and 235 g have been selected. (3 marks)

Solution

(7,0.7606)( 5) 0.7606 0.3064 0.7606 0.2330

W bin

P w

Specific behaviours states distribution and its parameters calculates ( 5)P w correctly

calculates final probability correctly

EXAMINATION 6 MATHEMATICS METHODS MARKING KEY CALCULATOR-ASSUMED Rebecca also sells oranges. The weights of these oranges are normally distributed. It is known that 5% of the oranges weigh less than 153 g while 12% of the oranges weigh more than 210 g. (d) Determine the mean and standard deviation of the weights of the oranges. (4 marks)

Solution 2( , )

(Z ) 0.05 1.6449P(Z ) 0.12 1.1750

153 2101.6449 1.1750

186.2 20.2

Z Norm

P z z

z z

xz

and

Specific behaviours calculates correct z-score for 0.05

calculates correct z-score for 0.12

generates simultaneous equations correctly solves correctly for and

EXAMINATION 13 MATHEMATICS METHODS MARKING KEY CALCULATOR-ASSUMED Question 17 (8 marks) A random sample of 100 people indicated that 19% had taken a plane flight in the last year. (a) Determine a 90% confidence interval for the proportion of the population that had taken

a plane flight in the last year. (2 marks)

Solution

Hence 0.125 ≤ p ≤ 0.255

Alternative solution

(1 ) (1 ) 100, 1.645, 0.19

0.19(1 0.19) 0.19(1 0.19)0.19 1.645 0.19 1.645100 100

0.125 0.255

p p p pp z p p z n z p

n n

p

p

Specific behaviours correctly calculates lower value of confidence interval correctly calculated upper value of confidence interval

EXAMINATION 14 MATHEMATICS METHODS MARKING KEY CALCULATOR-ASSUMED Assume the 19% sample proportion applies to the whole population.

(b) A new sample of 200 people was taken and X= the number of people who had taken a

plane flight in the last year was recorded. Give a range, using the 90% confidence

internal, within which you would expect X to lie. (1 mark)

Solution 100 0.125 100 0.254 13 25X X

Specific behaviours correctly calculates upper and lower value of interval

(c) Determine the probability that in a random sample of 120 people, the number who had

taken a plane flight in the last year was greater than 26. (3 marks)

Solution

The distribution is binomial with p = 0.19 and n = 120. ( 26) ( 27)P X P X , since n is discrete

Hence the required probability is 0.1928 (to four decimal places)

Specific behaviours

identifies the distribution as binomial – bin(120,0.19) uses 27 as the lower bound in the binomial cumulative distribution states the correct probability

(d) If seven surveys were taken and for each a 95% confidence interval for p was

calculated, determine the probability that at least four of the intervals included the true

value of p. (2 marks)

Solution

(7,0.95) (4 7) 0.9998bin P x

Specific behaviours identifies the distribution as binomial – bin(7, 0.95) calculates the probability correctly

EXAMINATION 15 MATHEMATICS METHODS MARKING KEY CALCULATOR-ASSUMED Question 18 (10 marks) A random survey was conducted to estimate the proportion of mobile phone users who favoured smart phones over standard phones. It was found that 283 out of 412 people surveyed preferred a smart phone.

(a) Determine the proportion p of those in the survey who preferred a smart phone.(1 mark)

Solution

283 0.6869412

p

Specific behaviours

calculates p correctly

(b) Use the survey results to estimate the standard deviation of p , for the sample

proportions. (2 marks)

Solution

283 283(1 )412 412 0.0228

412Standard deviation

Specific behaviours substitutes correctly into standard deviation formula calculates standard deviation correctly

EXAMINATION 16 MATHEMATICS METHODS MARKING KEY CALCULATOR-ASSUMED (c) A follow-up survey is to be conducted to confirm the results of the initial survey. Working

with a confidence interval of 95%, estimate the sample size necessary to ensure a margin of error of at most 4%. (3 marks)

Solution

283 283(1 )412 4120.04 1.96

516.366517

nn

n

Specific behaviours writes an equation to evaluate n from the margin of error

solves correctly for n

rounds n up to the nearest whole number

EXAMINATION 17 MATHEMATICS METHODS MARKING KEY CALCULATOR-ASSUMED

The 90% confidence interval of the sample proportion p from the initial survey is

0.649 p 0.725.

(d) Use the 90% confidence interval of the initial sample to compare the following samples: (i) A random sample of 365 people at a shopping centre found that 258 had a

preference for a smart phone. (2 marks)

Solution

258 0.71365

p and 0.668 p 0.746

The confidence interval for this second survey overlaps, significantly, the 90% confidence interval of the initial survey so this indicates we are sampling from the same population.

Specific behaviours

calculates 90% confidence interval for p correctly

states the similarity of results

(ii) A random sample of 78 people at a retirement village found that 32 had a

preference for a smart phone. (2 marks)

Solution

32 0.4178

p and 0.319 p 0.502

The confidence interval for this sample is quite different than that of the original survey. While this could be a random outlier it is more likely to be a biased survey from inside the retirement village.

Specific behaviours

calculates 90% confidence interval for p correctly

states the difference of results

EXAMINATION 23 MATHEMATICS METHODS MARKING KEY CALCULATOR-ASSUMED (e) Based on your observations of the graphs in this question, make a conjecture about the

defining rule for '( ).A x (1 mark)

Solution '( ) ( )A x f x

Specific behaviours states that '( ) ( )A x f x

Question 21 (5 marks) The graph below shows the number of faulty batteries per packet of 50 AAA batteries, when 50 packets are sampled at random.

(a) Identify the type of distribution of X = the number of faulty batteries per packet of 50

AAA batteries. (1 mark)

Solution The batteries tested are either faulty or not faulty. Each test of a battery is a Bernoulli trial. Hence the underlying distribution is binomial.

Specific behaviours identifies the distribution as binomial

EXAMINATION 24 MATHEMATICS METHODS MARKING KEY CALCULATOR-ASSUMED A manufacturer of AAA batteries assumes that 99% of the batteries produced are fault-free. Ten samples of 50 packets of 50 AAA batteries are selected at random and tested. The number of faulty batteries in each of the 10 random samples is provided below.

Sample 1 2 3 4 5 6 7 8 9 10

Number of faulty batteries

34 28 22 38 28 30 22 16 28 30

(b) Using the assumption that 99% of batteries are fault free calculate the 95%

confidence interval for the proportion of faulty batteries expected when sampling. (3 marks) (c) Decide which of the samples, if any lie outside the 95% confidence interval.

(1 mark)

Solution

Using 25 out of 2500 batteries being faulty. This gives us 0.01p and a 95%

confidence interval of:. . 0 0060997 0 0139003 35p n n 15 where the number of faulty batteries

Sample 4 n=38 lies outside the interval.

Specific behaviours

identifies n = 2500 and x = 25 as the required variables to calculate the 95%

confidence interval determines the 95% confidence interval calculates the interval of faulty batteries determines the sample which lies outside the 95% confidence interval

CAS Use binomcdf - · PDF fileOnly worth 1 mark so fine to use the 1-prop z interval function...

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