stpm term 3 rev test 3

8/11/2019 STPM Term 3 Rev Test 3

1/23


2/23

!or 1 x

i"nore limits

$1

orrect inte"ration correct limits seen or used

*1

addin" or!(1)

m1

*+ (*)

*1

Alternative

($1)

!(1) % c% 1

(m1)(*1)

!(x) % 1 (x- )


3/23

(*1)

Alternative

(c) /(2 X ) %

!() - !(2)

$1

*1

2

(d) (i)

use of !(q) %

$1


4/23

(either)

$1

(0) (q- )% -1.

AG

*1

(ii)

*+

&1

1

12

* machine fills small cans with soda water. The volume of soda water delivered by the machinemay be modelled by a normal random variable with a mean of 1 ml and a standard deviationof 1.3 ml.

4ach can is able to hold a ma5imum of 1 ml of soda water.

/rinted on each can is 6ontents 17 ml8.

(a) 9etermine the probability that the volume of soda water delivered by the machine:

(i) does not cause a can to overflow;

(ii) is less than that printed on a can.

(3)

(b) !ollowin" ads of 12 cans may be assumed to be random samples of cans filled by the


5/23

machine, determine the probability that, in a pac>, the mean volume of soda water per can ismore than 12. ml.

()

(Total 17 mar>s)

me,X? @(1, 1.32)

(a) (i) /(not overflow) % /(XA 1)

$ay be implied

&1

Standardisin" (1., 1 or 1. or1B., 17 or 17.) with 1 and

( , 1.3 or 1.32) andCor (1 - 5)May be gained in (a)(i) or (a)(ii)

$1

% /(ZA 1.25)

*+; i"nore inequality and si"n$ay be implied by a correct answer

*1

% 0.894 to 0.895

*D!D (7.=B)

*1

(ii) /(less than printed) % /(XA 17)

+nly if &1 not awarded in (a)(i)

(&1)

% /(ZA -1.=E)1 - /(ZA 1.=E)

*rea chan"e$ay be implied by a correct answer


6/23

or answer A 7.

$1

% 0.03 to 0.031

*D!D (7.777)

*1

3

(b) Folume, Y? @(12, 7.=2)

Fariance of % 0.82/12 0.053

Sd of % 0.8/G12 0.23 to 0.231

*+C*DHTStated or used*+C*D!D

&1

Standardisin" 12. with 12 or 1 and

or or equivalent; allow

(12 - )

$1

% /(ZI 2.13)% 1 - /(

ZA 2.13)

*rea chan"e$ay be implied by a correct answeror answer A 7.

m1

% 0.015 to 0.0153

*D!D (7.711B)

(1 - answer) &1 $1 ma5


7/23

*1

17J

The number of telephone calls received, during an 8-hour period, by an IT company thatreuest an urgent visit by an engineer may be modelled by a !oisson distribution with a meanof ".#a$ %etermine the probability that, during a given 8-hour period, the number of telephone calls

received that reuest an urgent visit by an engineer is&

(i) at most ;(1)

(ii) e5actly E;

(2)

(iii) at least but fewer than 17.

()

(b) Drite down the distribution for the number of telephone calls received each hour thatrequest an ur"ent visit by an en"ineer.

(1)

(c) The KT company has en"ineers available for ur"ent visits and it may be assumed thateach of these en"ineers ta>es e5actly 1 hour for each such visit.

*t 17 am on a particular day, all en"ineers are available for ur"ent visits.

(i) State the ma5imum possible number of telephone calls received between 17 am and 11am that request an ur"ent visit and for which an en"ineer is immediately available.

(1)

(ii) alculate the probability that at 11 am an en"ineer will notbe immediately available toma>e an ur"ent visit.

()

(d) ive a reason why a /oisson distribution may not be a suitable model for the number oftelephone calls per hour received by the KT company that request an ur"ent visit by an en"ineer.


8/23

(1)

(Total 1 mar>s)

(i) X? /o(E)/(X ) % 7.71

*D!D 7.77 and 7.71

&1

1

(ii)

$1

% 7.1B

*1

/(L E) - /(L 3)% 7.B=E - 7.BE ($1)% 7.1B (*1)

2

(iii) 7.3 p 7.33

/(L B) - /(L )

&

7.E2 p 7.E or 7.2p 7.

/(L 17) - /(L )/(L B) - /(L )

(&2)

7.37

/(L 17) - /(L )

(&1)


9/23

(b) @o. telephone calls received per hour% Y? /7(7.=E)

&1

1

(c) (i) $a5imum number %

&1

1

(ii) /(YA ) % /(Y% 7, 1, 2, )

% 7.13B(1 7.=E 7.=2= 7.111E)

% 7.B=EE7

*ny correct e5pression (&2)or *D!D 7.B=E to 7.B==

&2

/(YM ) % 1 - 7.B=EE

1 - (their /(#A ))

$1

% 7.712

*D!D 7.7122 and 7.712

*1

!"

/(# ) % 7.BBE to 7.BB=or any correct e5pression

&2

/(#I ) % 7.772 to 7.77 $1 *7

(d) probably not constant


10/23

The number of calls in any time intervalof 1 hour is li>ely to vary throu"hout the day.

6System 9own8not independent

41

1

1J

* district council claimed that more than =7 per cent of the complaints that it received about thedelivery of its services were answered to the satisfaction of complainants before reachin" formal

status.

*n analysis of a random sample of 1E complaints revealed that 2= reached formal status.

(i) onstruct an appro5imate BN confidence interval for the proportion of complaints thatreach formal status.

()

(ii) Oence comment on the council8s claim.

(2)

(b) The district council also claimed that less than 7 per cent of all formal complaints weredue to a failin" in the delivery of its services.

*n analysis of the 7 formal complaints received durin" 277EC7= showed that 13 were due to afailin" in the delivery of its services.

(i) Psin" an e5act test, investi"ate the council8s claim at the 17N level of si"nificance. The7 formal complaints received durin" 277EC7= may be assumed to be a random sample.

()

(ii) 9etermine the critical value for your test in part (b)(i).

(2)

(iii) Kn fact, only 2 per cent of all formal complaints were due to a failin" in the delivery ofthe council8s services.

9etermine the probability of a Type KK error for a test of the council8s claim at the 17N level of

si"nificance and based on the analysis of a random sample of 7 formal complaints.


11/23

()

(Total 1= mar>s

(i) 0.1#

*+; or equivalent

&1

BN z% 1.9#

*DHT

&1

*ppro5imate K forpis

Psed

$1

ie 7.13 Q 1.B3

+r equivalent

! on and R

*1!

ie 0.1# Q 0.054 or (0.10#$ 0.214)

*+C*DHT or *DHT (7.7)*1

(ii) "%does include 0.2(27N)

! on (i)

&1!

&o eviden'e to suortcouncils8 claim


12/23

! on (i)9ependent on K and on 7.2

&1!dep

2

(b) (i) O7:p% 7.7 (7N)O1:pA 7.7

&oth

&1

Psin" & (7, 7.) (7N)

$ay be implied

$1

/(X 13) % 0.15#

*DHT (7.131)

*1

alculated probability I 7.17 (17N)

omparison

$1

&o eviden'e, at 17N level, to suortcouncil8s claim!e'ial "ase@ormal appro5imationz%*1.15(E) &1 F %*1.28(13) &1onclusion &1! $a5 of mar>s

! on probability v7.17 or 7.7*t 17N level, a''et (at least) 40+*llow &1 for hypotheses

p% 7.12 to 7.12 v7.17 &1 &1! on Rand F

*1!

(ii) Hequire /(Xx) 7.17


13/23

$ay be impliedK"nore any reasonin" if 618 stated

$1

F % 15 (H 1)

*+; or equivalent

*1

2

(iii) /(Type KK error) % /(accept O7 O7false)

Stated or used; or equivalent

&1

% /(XI F orXM F)

*ttempt at a probability I or M 8s F from (ii)

$1

% 1 - (0.83#9 or 0.,481)

K"nore 61 -8

$1

% 0.1#3

*DHT

*1

1=

(a) * district council claimed that more than =7 per cent of the complaints that it receivedabout the delivery of its services were answered to the satisfaction of complainants beforereachin" formal status.

*n analysis of a random sample of 1E complaints revealed that 2= reached formal status.

(i) onstruct an appro5imate BN confidence interval for the proportion of complaints that


14/23

reach formal status.

()

(ii) Oence comment on the council8s claim.

(2)

(b) The district council also claimed that less than 7 per cent of all formal complaints weredue to a failin" in the delivery of its services.

*n analysis of the 7 formal complaints received durin" 277EC7= showed that 13 were due to afailin" in the delivery of its services.

(i) Psin" an e5act test, investi"ate the council8s claim at the 17N level of si"nificance. The7 formal complaints received durin" 277EC7= may be assumed to be a random sample.

()

(ii) 9etermine the critical value for your test in part (b)(i).

(2)

(iii) Kn fact, only 2 per cent of all formal complaints were due to a failin" in the delivery ofthe council8s services.

9etermine the probability of a Type KK error for a test of the council8s claim at the 17N level ofsi"nificance and based on the analysis of a random sample of 7 formal complaints.

()

(Total 1= mar>s)

(a) (i) 0.1#

*+; or equivalent

&1

BN z% 1.9#

*DHT

&1

*ppro5imate K forpis


15/23

Psed

$1

ie 7.13 Q 1.B3

+r equivalent

! on and R

*1!

ie 0.1# Q 0.054 or (0.10#$ 0.214)

*+C*DHT or *DHT (7.7)

*1

(ii) "%does include 0.2(27N)

! on (i)

&1!

&o eviden'e to suortcouncils8 claim

! on (i)9ependent on K and on 7.2

&1!dep

2

(b) (i) O7:p% 7.7 (7N)

O1

:pA 7.7

&oth

&1

Psin" & (7, 7.) (7N)

$ay be implied

$1

/(X 13) % 0.15#


16/23

*DHT (7.131)

*1

alculated probability I 7.17 (17N)

omparison

$1

&o eviden'e, at 17N level, to suortcouncil8s claim!e'ial "ase@ormal appro5imationz%*1.15(E) &1 F %*1.28(13) &1onclusion &1! $a5 of mar>s

! on probability v7.17 or 7.7*t 17N level, a''et (at least) 40+*llow &1 for hypotheses

p% 7.12 to 7.12 v7.17 &1 &1! on Rand F

*1!

(ii) Hequire /(Xx) 7.17

$ay be impliedK"nore any reasonin" if 618 stated

$1

F % 15 (H 1)

*+; or equivalent

*1

2

(iii) /(Type KK error) % /(accept O7 O7false)

Stated or used; or equivalent

&1

% /(XI F orXM F)


17/23

*ttempt at a probability I or M 8s F from (ii)

$1

% 1 - (0.83#9 or 0.,481)

K"nore 61 -8

$1

% 0.1#3

*DHT

*1

The random variableXhas a probability density function

!or this distribution, you are "iven that

4(X) % and 4(X2) % 22- =

(a) !ind, in terms of , the variance ofX.

(2)

(b) The mean of a random sample of nobservations,X1,X2,X, ...,Xn, is denoted by .

(i) Drite down, in terms of and n, e5pressions for the mean and the variance of .

(2)

(ii) 45plain why is an unbiased and consistent estimator for .

(2)

(c) (i) !or a random sample of siRe , the median,M, is an unbiased estimator for

with variance 2- .


18/23

!or such a sample, calculate the relative efficiency ofMwith respect to , "ivin" your answerto three decimal places.

Oence "ive a reason why should be preferred toMas an estimator for .

()

(ii) * particular random sample of observations ofX"ave the followin" values.

7.=7 1.BB .12 .=B 3.27

(A) iven that /(XM 2) % 7, use this sample to find an inequality for .

(-) +btain the values of and mfor this sample.

(") omment on your answers in part (c)(ii)(&) in the li"ht of your answers to part (c)(i).

()

(Total 1 mar>s

(a) Far(X) % 2U - = - U % U - =

$1*1

2

(b) (i) 4( ) %

&1

&1

2

(ii) 4( ) % unbiased

41

Far( ) V 7 as nV W consistent

41

2


19/23

(c) (i)

*ny sensible value for

$1

% 7.3 or 7.33

*1

/refer since H4(Mwrt ) A 1

or Far( ) A Far($)

41

(ii) (A) 2> 3.2 I .1

M is $1*7

$1*1

(-) % .27 m% .12

both

&1

(") is the more efficient estimator,implyin" that for the ma


20/23

&efore leavin" for a tour of the PX durin" the summer of 277=, 4duardo was told that the PXprice of a 1.Ylitre bottle of sprin" water was about 7p.

Dhilst on his tour, 4duardo noted the prices,xpence, which he paid for 1.Ylitre bottles ofsprin" water from 12 retail outlets.

Oe then subtracted 7p from each price and his resultin" differences, in pence, were

-1= -11 1 1 E -1 1E -13 1= - 7 B

(a) (i) alculate the mean and the standard deviation of these differences.

(2)

(ii) Oence calculate the mean and the standard deviation of the prices,xpence, paid by4duardo.

(2)

(b) &ased on an e5chan"e rate of Z1.22 to [1, calculate, in euros, the mean and the standarddeviation of the prices paid by 4duardo.

()

(Total E mar>s

-1= -11 1 1 E -1 1E -13 1= - 7 B

(i) $ean, % 1.5

*+ \d% 1=K"nore notation and units (11.EE or 12.2B)

Standard deviation,dorsd% 11., to 12.3

*D!D \d2% 13=7

$ean, % 7 % 51.5

! on (a)(i) or correct


21/23

&1!

2 B 1 3 E B 3E 3= E 7 B

% 31= \52% =7

K"nore notation and units

Standard deviation, xorsx% 11., to 12.3

! on (a)(i) providin" I 7 or correct

&1!

'(alues, mean or sd in #a$#i$ or #a$#ii$) *Award if use seen or implied by + 1Subseuent correct or #correct * 1$answer

ean 0.628 to 0.63A/0/ #.283$

Standard deviation 0.14 to 0.151A/0/ #.1432 or .145$

Special Cases:

At least one answer correct with no statedunits or incorrect stated units 1 A1A1 ma6At least one answer * 1 with its units

stated as 7cents 1 A1 A1 ma6At least one answer * 1 with no unitsstated or units stated as euros 9 pence 9 :

1 onlycents attached to + 1 answer * 1


22/23


23/23

24 12 1>

< 12 88 1

S 12 8 8

/ >" 113 1"

Total 1> 35> >

D & Selection independent of home region

D1 &

Selection not independent of Dome region1 tail 1E

2

Cal 24.5" F 11.34> Ce@ect Do

c$ #i$ There is significant evidence to suggest that selection is not independent of home

region.#ii$ Artists from the south seem less li=ely to be selected # e6pected higher than observed$and those from the west seem much more li=ely to be selected #e6pected lower thanobserved$.

stpm term 3 rev test 3

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