2020_stas_Alevel_nov_paper1

Extracted Content: 2020_stas_Alevel_nov_paper1

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ZIMBABWE SCHOOL EXAMINATIONS COUNCIL

General Certificate of Education Advanced Level

MARKING SCHEME

NOVEMBER 2020 SESSION

STATISTICS ~ 6046/1

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2 | 6 8 PO = 99+9O+O0@ MI = 18] 01 SSEB Al 120 21 (b) P(E) = OG) BiBildds Ll - 5 lo. cs34 Al 6) 2 (a) Arranging numbers in order of size:

41 42 42 42 43 44 45 46 46 47 48 55

Q. = 44.5 Bl
Qi = 42 and Q3 = 46.5 BIBI
FA
(b) On graph paper ecake gt ler
(c) Distribution nearly symmetrical I normnes \ Bl YW é)
3 (a) (i) Number of ways of choosing team = C31
= aul
- 4!7!
= 330 Bl

(ii) | Number of ways of choosing 3 men and one woman

= (C$)(C¥)
- GG)
= 100 Al

(b) | When one man is chosen, then 3 women must be chosen
n(M) = (€?)(c8)
= St) (s!
(S) (5) Mt
= 60 Ar Bho (6) j

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..P(M)

E(X) = 3.5

1(0,1) + 2(0,3) + a(0,4) + b(0,2) = 3,5

Qa+b=14

Var(X) = 2.65

1? (0,1) + 27(0,3) + a (0,4) +b? (0,2) — 3.5? = 2.65

(a)

(b)

(a)

(b)

2a° +b’ = 68

2a’ + (14 - 2a)’ = 68

(3a —16)(a—4) =0

4(a= 5: not admissible)

b = 6

Dix = 225, Yx? = 8875, Ny = 361,

Dy? = 22641, Yxy = 12905, n=6

_ 6(12905)~(22)(361)
~~ 6(8875)—(225)2

= — -1.446
y — 60.17 = —1.446(x — 37.5)
= y =-145x +1144

No, because x has been controlled

ga 2a 414 [ul 2eea
2 — 33951 _ 44 ge

18

AZ. \2&
£72-2066667

(i) Yx = 41X17 = 697
Mass of pupil who left = 745 — 697

= 48 kg

60 2
ou MIAI G

Bk
(a) 8 |

M1
(2) ~Al

MiaHempt to She

Ciprul Kaan cosy

Al

Al
[6]

MI come ch sole}
Al

MIAI

BIBI {2}
[6]

Bl

MI

Al 4]

MI

Al

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(a)

(b)

(c)

(a)

(b)

4

(ii) S _ see 412 = 13.4 MIAI BR}

[7]

PX =2)= §(2)’ yey “
Ora
- $a)
= 200036786 b22 Al
o.oo | age UE
P(X2 4)=1-(PX <4) | Pca oS) Bl

s

- -aP-aQ Qt eg G-40 0
= 1-(9)°-6 (9-15-20) (9)

= 1-S0TS980TT
= Gee0876T987~ o.coast | 17 w
P(X < 3) = P(X = 0) + P(X = 1) + P(X =2)

FG)"+ oH MI
- JOO +s Q

ae ks

= 093743347 O° Al Tat
[7]

P(X = 4) = (0,6)(0,4) M1

= — 0,0384 Al i
BX) =

_ oe)

= §|1.b BI
VatX)= oes MI

» FE 1s
- ® “a NMI Al

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5

(c) P(X 23)=P(X =3) + P(X > 3) [eOe>2) BI
= eooayrroa oF Ml
= 1G Al
0,16 © i
9 (a) C [g(x — x2) dx = ¢ fx? -2F Ml
= [= _ =| Ml
4 24
= 28€ Al {3]
= a >0
. 25c M1
(b) (i) m 1
= 24 Al 2
Cc = _ [ ]
Gi) EX) = = fe(2x? — x3) dx
= 24 [2x3 “yp M1
7 2513 4]
247125 625
- e-F m
= 3 oS Al 3]
2 lob
S | ° [8]
10 (a) PQ<X< 5) = (= < > < >) Ml glundadisHe
1 2
= P(-}<2<?)
= —_ 0,7477 + 0,6304 - 1 M1
= 037% OSs | Al
(b) P(X>0) = P(Z> SS) MI
= P(Z>-1)
= (1)

= —0,8413 Al

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12

(c)

(a)
(b)

(c)

(d)

(a)

6
P(X — 3| > 6|= P(X <3) + P(X > 9)
= p(z<==)+P(z>)
= P(Z<-2) + P(Z>2)
= 21-2]
= 21 -0,9772]

= 0,0456

Number of years used = 8
Sales for 1985:
J = 284+ 14.4 (11)

= Se YEA 4 (ugar. Woo
Annual increase in sales = 14.4 fin (OND

n~ 284 14.4 x
12 12 12

= 23,67 + 0,1x

Binley kato

M1

Ad B\ vabale volt

Al a

Bl

MI
Al

Bl

MI

To move origin to 15 July 1980, add 36,5 increments of 0,1 to

23,67:

J = 23,67 + 36.5(0,1) + 0,1x
= 27.32 + 0, 1x

Sales for September 1980:

J = 27.32 + 0,1(2,5)

= 2757
P(X < 20) = 10-9 =
4720
= ot FA.
1 1
- fF -3]
- 2
~ 2

M1

Al

M1

Al

Ml

Al D&

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(b)

(a)

(b)

(c)

Forx> 10, F(x) = 10 f* at
0 ¢2

~ wT,

10
3-20
x

i]

For x < 10, F(x) =0

4-22
“Fe ={ Z*% > 10
0 1x <10

Y ~ Bin (6,3)

P(Y 2 3)=1-P(Y <3)

- 1-6() 6) -e@E

= 1 -0,34375

= 0,65625

P(A > 65) =P(Z > =")

= P(Z>-1.667)

= 0,9522

M=A,+ Az + A3— (Bi + Bz + Bs) ~ N(B0,183)

0-30
P(M > 0) = rz ? a)
= P(Z>-2.218)
= —0,9867

H=A-—1.15 B~N(0.25, 69.0625)

P(H > 0) = P(z > oo)
= P(Z > —0,03008)
= (0,03008)
= 0,5120

y _
= 1-('-66'-56)

M1

Al

Bl va
Bl

Bl

Ml

Al

<s

MI

Al

BIBI

M1

Al

BIB1

MI

Al
[10]

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(a) X ~ Bin(500,0.4)

np = 200 > 5 and nq = 300>5

X ~N(200,120) BIBI
P(X = 190) = P(189.5 <X < 190.5) Bl
_ 189.5—200 190,5-200 M1

P ( ¥120 <Z< 120 )

= P(-0,9585 < Z <—0,8672)
= (0.9585) - (0.8672)

= 0,8312 — 0.8070

= 0,0242 Al
(b) — P(L80< X < 210) = P(179.5 < X < 210.5) Bl
_ 179.5-200 210.5—200
= P( ae <2 < ee ) MI

= — PC-1.871 <Z < 0.9585)
= — @(1.871) + (0.9585) - 1

= 0,9694 + 0.8312 - 1

= 0,8006 Al
(c) P(X> 180) = PCK> 180.5) BI

= P(z >See) MI

= P(Z>-1.780)

= (1.780)

= 0,9625 Al

[11]

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9 Ho: Sample population has Poisson distribution BI eae Ha: Sample population has some other distribution a= 0,01 df=8-1-0=7 Bloor Reject HO if X21 > 18,48 Bl Ey = 600x2— 2 = 49,2 MIAI MIAI Bi Ei (O-EV/E 34 49.3 yq.29| 4.723 Of .T}226 131 123 0.504 160 153.9 0.242 136 128.3 0.467 72 80.2 0.838 37 40.1 0.240 22 16.7 1.682 8 8.5 0.029 Al1Al1 Xe, = 8.725 MIAI

Since X2,, = 8.725 < 18.48, we do not reject HO and conclude that V7 LAH
the sample population has a Poisson distribution. [13]

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