2022_PAPER_1_SOLUTION_FOR_TEST_1

Extracted Content: 2022_PAPER_1_SOLUTION_FOR_TEST_1

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iy ONLINE REVISION

General Certificate of Education Advanced Level

STATISTICS 6046/1
PAPER 1 SOLUTION
REVISION TEST 1 2022 SESSION 3 hours
Additional materials:
Answer paper
Graph paper

List of formulae MF7
Electronic calculator (Non-programmable)

Time 3 hours

INSTRUCTIONS TO CANDIDATES

Write your name in the spaces provided on the answer sheet/answer booklet.

Answer all questions.

If anumerical answer cannot be given exactly, and the accuracy required is not specified in the
question, then in the case of an angle it should be given correct to the nearest degree, and in other
cases it should be given to 2 significant figures.

INFORMATION TO CANDIDATES

The number of marks is given in brackets [ | at the end of each question or part of question.
The total number of marks for this paper is 120.

The use of a scientific calculator is expected, where appropriate.

You are reminded of the need for clear presentation in your answers.

This question paper consists of 7 printed pages and 1 blank page.

Copyright: Sharewell B. C. (Share)—Share Online Revision Paper 1 2022

© Sharewell B. C. (Share)—Share Online Revision Paper 1 2022
Complied and typed by Share + 263784089807 [Turn Over

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1.

In a computer game played by a single player, the player has to find, within a fixed
time, the path through a maze shown on the computer screen. On the first occasion
that a particular player plays the game, the computer shows a simple maze, and the
probability that the player succeeds in finding the path in the time allowed is 0.75.
On subsequent occasions, the maze shown depends on the previous game. If the
player succeeded on the previous occasion, the next maze is harder, and the
probability that the player succeeds is one half of the probability of success on the
previous occasion. If the player failed on the previous occasion, a simple maze is

shown and the probability of the player succeeding is again 0.75.

The player plays three games.

a) Show that the probability that the player succeeds in all three games is =. [4]
b) Find the probability that the player succeeds in exactly one of the games. [3]
c) Find the probability that the player does not have two consecutive successes. [3]
d) Find the conditional probability that the player has two consecutive success

given that the player has exactly two successes. [3]

SOLUTION

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1% game 274 game 374 game
oars . 0.8125 5!
7 H

a) P(player succeeds in all three games ) = P(SSS)
= (0.75 x 0.375 x 0.1875)

27
=sn (shown)

b) P(player succeeds in exactly one of the games) = P(SSS) + P(SSS) + P(SSS)
= {(0.75 x 0.625 x 0.25) + (0.25 x 0.75 Xx 0.625) + (0.25 x 0.25 x 0.75)}

9

32
= 0.28(2s.f)

c) P(player does not have two consecutive successes) = P(SSS)

= 1 -(0.75 x 0.625 x 0.75)

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_ 83
~ 128

= 0.64 (2s.f)

d) P(two consecutive success given that the player has exactly two successes)
P(SSS)+P(SSS)
P(SSS)+P(SSS)+P(SSS)

7 (0.75x0.375x0.1875)+(0.25x0.75x0.375)
~ (0.75x0.375x0.1875)+(0.75x0.625x0.75)+(0.25x0.75x0.375)

P(2S | 2S) =

17
37

= 0.46(2s.f)

2. The amounts of money, x dollars, that 24 people had in their pockets are

summarized by (x — 36) = —60 and )\(x — 36)? = 227.76. Find }x and yx”. [5]

SOLUTION
¥(x — 36) = -60
yx —¥36 = —60

1 ¥ix = 24 x 36 — 60
yx = 804
Now, ¥\(x — 36)? = 227.76
¥(x2 — 72x + 1296) = 227.76
Yix2 — 72x + 1296 x 24 = 227.76
w ¥x? = 72 x 804 — 1296 x 24 + 227.76

yx?=27011.76

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3. Acode consists of 10 digits, 4 of which are zeros and 6 of which are ones. Calculate

the number of such blocks in which the first and last digits are of the same as each

other. [4]
SOLUTION

0000111111

When the first and last digit is 0, then we treat 2 0s as non - existing

-. possible number of blocks = — = 28

When the first and last digit is 1, then we treat 2 1s as non - existing

-. possible number of blocks = =~ = 70

Now, total possible number blocks = (28 + 70) = 98 ways

a) An unbiased tetrahedral die has faces marked 1, 2, 3, 4. If the die lands on the
face marked 1, the player has to play 10 USD.
If it lands on a face marked with a 2 or a 4, the player wins 5 USD and if it
lands on a 3, the player wins 3 USD. Find the expected gain in one throw. [5]
b) A discrete random variable X can take values 10 and 20 only. If E(X)= 16,
write out the probability distribution of X. [7]

SOLUTION

a) Let x be the amount paid or gained on each throw

x -10 5 3

1 1

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BIR

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Expected gain; (—10 xz) + (5 x5) x (3 x =) =+ = USD 0.75

b) x 10 20

P(X =x) a b

Now, (p.d. fy =1

SAPD=1L oan (1)

Since, E(X)=16

Then, 10a + 20b = 16 we, (2)
a=1-b

10(1 — b) + 20b = 16

10 —10b + 20b = 16

10b=6

eaat [Substitution of equation (1) ]

Now, p.d.f of X is;

x 10 20
4 6
P= x) 10 10

RES: MUTARE

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5. The continuous random variable H has continuous p.d.f. f(h) where

h 2
373 2<5hs3
fh) =4 a 3<shs5
2—Bh 5<h<6
0 otherwise
Find
a) aandp [2]
b) F(h) and sketch y = F(h) [4]
c) P(2<H <3.5) [2]
SOLUTION
a) s-iso and o=2-£h
3 2 to5_
5-5 =a += 2-66)
1
o= 3 56 =-
p=
nz 2<sh<3
3 3
1
= 3<h<5
b) f(A) =4 3 ~
2-=h 5 <h<6
0 otherwise
t(h 2 _ 1\h? 2 t
1Row; f) (=-=) ah = (@£-24)]5
2 t? 2
=[-3-(5-79)
2 tv 2 .
=-t--—-- [A sketch of this graph, shows that the area
3 6 3 >

is below the x-axis yet area cannot be

for instance (-A units?) ]

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2 tt 2 wt 2 2
Therefore from 1°' Row we get; -[2e-+-3] =—+---t
3 6 3 6 3 3
2nd Row; J, =ah = [=n] 5
173 3 3°13
3 421
= [> -=¢| [A sketch of this graph, shows that the area

is below the x-axis yet area cannot be

for instance (-A units?) ]

Therefore from 2 Row we get; — [> — =e] =-1 +it

3rd Row; (2 - =n) dh = [2n 7 +) 5
== (e-)
_ [=¢? _ at +=] [A sketch of this graph from 5 to 6,

shows that the area
is below the x-axis yet area cannot be

for instance (-A units?) ]

5 5
Therefore from 3'¢ Row we get; — [=¢? - 2t +=] = -i0? -k 2t— =

Now, l' Row, then 1%' Row + 2 Row, then finally 1°* Row + 2™ Row + 3° Row
Will give us the cumulative distribution function, F(A).

In general, cumulative means incorporating all data up to the last!

Therefore,

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2g? 2<hs3
3°63
+4 +( 1+5h) 3<A<5
3 63 3
F(A) =
34 042)4(-144@ +(-Z1 42h =) S<h<6
63 3 6
1 h>6
2 ey? 2<hs3
3.6 3
5 1
F(h) = atah 3<h<5
—sh? +2h—5 <h<6
1 h

c) Ans. (= 5) [integrate]

6. Arandom sample of St Dominic’s mathematics students marks for end of term test

are shown below.

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67 76 85 42 93 48 93 46 52
72 77 53 41 48 86 78 56 80
70 70 66 62 54 85 60 58 43
58 74 44 52 74 52 82 78 47
66 50 67 87 78 86 94 63 72
63 44 47 57 68 81

a) Contract a stem a leaf diagram to represent these data. [6]
b) Find the median and the quartiles for this distribution. [6]
c) Draw a box plot to represent these data. [4]
d) Give one advantage of using

i) a stem and leaf diagram

ii) a box plot,

to illustrate data such as that given above. [2]

SOLUTION

a) stem leaf
1234467788
0222346788
023366778
00224467888
01255667
334

ow ON DD HN Bs

b) Q, = 66 miles, Q, = 52 miles, Q, = 78 miles

cc) ——L_] t+} —

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40 50 =660 70 80 90 [distance (miles)]

d) Keeps original data

Shows how symmetrical the data is.

7, Calculate the equation of the regression line x on y for the following distribution,

x 22 36 25 41 35 40
y 78 70 65 58 48 42

and the product-moment correlation coefficient, (7) and comment your value. —_ [7]
SOLUTION

The equation of the line required can be written in the form;

x=at+by where a =—by and b =.
Syy

So, the equation of the regression line X on Y is; (x -—*) =b(y—y)

_ nyxy-DxDy
nyy?—Oy?

b= 6(11599)-(199)(361)
~~ 6(22641)-(361)2

449
1105

b = —0.4063

(x — 33.16666667) = - 0 — 60.16666667)

x = — 4? y +:57.61447964
1105

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x= —0.4063y + 57.6145
Now, the product-moment correlation coefficient, (7) is;

_ nyxy — Uxyy
ynyx? — (dix)? x ynyy? — iy)?

6(11599) — (199)(361)

J6(6911) — (199)? x ,/6(22641) — (361)2

r = —0.6993760986
r = —0.70 (2s.f)

There is a strong negative correlation between the x and y values.

a) Inacertain country, 68% of households have a printer. Find the probability

that, in a random sample of 8 households, 5, 6 or 7 households have a

printer.. [4]

b) Use an approximation to find the probability that, in a random sample of 500

households, more than 337 households have a printer. [4]

c) Justify your use of the approximation in part (b). [1]
SOLUTION

a) Let X be the households with printers, with parameter p = 0.68 andn = 8
Then, X~Bin(8, 0.68)
Now, P(5, 6 or 7) = 8C(0.68)5(0.32)? + 8C(0.68)°(0.32)? + 8C(0.68)7 (0.32)!

= 0.722

b) Since, np = 500(0.68) = 340 > 5 and nq = 500(0.32) = 160 > 5

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Then, X~N(np,npq)
X~N(340, 108.8)
P(X > 337) will be P(X > 337.5) [using continuity correction]

P(X > 337.5) =P (z > ———)

108.8
= P(z > —0.2396)
= @(0.2396) [From tables]

= 0.595
c) Since, np = 500(0.68) = 340 > 5 and nq = 500(0.32) = 160 >5

9. Ina golf tournament, the number of times in a day that a ‘hole-in-one’ is scored is
denoted by the variable X, which has a Poisson distribution with mean 0.15. Mr
Kuna offers to pay $200 each time that a hole-in-one is scored during 5 days of play.

Find the expectation and variance of the amount that Mr Kuna pays. [4]
SOLUTION

X~Po(0.15) ina day.
So, for 5 days X~Po(5 x 0.15)...
Then the expectation of the amount that Mr Kuna pays is;

(0.75 x 200) = $150

And the variance is;

(2002 x 0.75) = $ 30000

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10. In the past, the flight time, in hours, for a particular flight has had mean 6.20 and
standard deviation 0.80. Some new regulations are introduced. In order to test
whether these new regulations have had any effect upon flight times, the mean flight

time for a random sample of 40 of these flights is found.

a) State what is meant by a Type! error in this context. [2]
b) The mean time for the sample of 40 flights is found to be 5.98 hours.
Assuming that the standard deviation of flight times is still 0.80 hours, test at
the 5% significance level whether the population mean flight time has
changed. [4]
c) State, with a reason, which of the errors, Type I or Type II, might have been
made in your answer to part (b). [2]
SOLUTION

a) Conclude flight times affected when in fact they have not been.

b) Let X be the flight time

Let the population mean be yw and the population standard deviation be o.
Ho} H = 6.2
Ho; w # 6.2

So, if Hg is true, then X~M(6.2, 0.807)

95%

Use a two tailed test, at 5% level of significance.

So, we reject Ho if Z.q; < —1.96 or Z,q; > 1.96

>i

TH
Now, Zeal al

al

5.98-6.2
* Lea = aa
a0

= —1.739252713 -196 -1.74 0 1.96

Since Z,q; > —1.96, there is no evidence that flight times are affected at 5% level

of significance.

c) Hp was not rejected, hence type II error.

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11.
a) A petrol station finds that its daily sales, in litres, are normally distributed
with mean 4520 and standard deviation 560.
i) Find on how many days of the year (365 days) the daily sales can be
expected to exceed 3900 litres. [4]
The daily sales at another petrol station are X litres, where X is normally
distributed with mean m and standard deviation 560. It is given that
P(X > 8000) = 0.122.
ii) Find the value of m. [3]
iii) Find the probability that daily sales at this petrol station exceed 8000
litres on fewer than 2 of 6 randomly chosen days. [3]
b) The random variable Y is normally distributed with mean yp and standard
deviation o. Given that 0 = = u, find the probability that a random value of Y
is less than 2p. [3]
SOLUTION
a)
i. X~N(4520, 5207) [let X be the daily sales of a petrol station]
3900-4520
P(X > 3900) = P(z > =")
P(z > —1.107)
= 6(1.107)
= 0.8657

Therefore number of days is; 365 x 0.8657 = 315 days.

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ii Since, P(X > 8000) = 0.122

m-—8000
560

Then, P (z > ) = 0.122

“Z= 1.165

8000—m
560

m = 7250

1.165 =

iii, Now, X~B(6, 0.122)
Therefore, P(X < 2) = (0.878)® + $C(0.122)1(0.878)°
= 0.84 (2s.f)

b) Y~N(w0)

. _2 7 2
Given that o = zlh then Y~N (u, 3 1)

por <2) =P (2 <4)

Su
= P(z < 1.5)
= 0.933

12. A firm of solicitors claims that, on average, interviews with clients last 50 minutes. A
random sample of 15 interviews is chosen, and the time taken for each interview, x
minutes, is noted. The results are summarized by }'.x = 746 and Y'x? = 37 180.
Assuming that the time for an interview has a normal distribution, determine at 5%

significance level, whether the firm is overstating the average interview. [6]
SOLUTION

Let X be the time taken on interviews with clients.

Let the population mean be yp, and the population standard deviation be o.

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x=

TEST 1 SOLUTION

= = 49,73333333

Dx
n 1

(dx 2 - 2")

crasy" =“)

“ty 180 —
~ ts 1
2°? = 5.638095238

~ 105

Ho; # = 50

Ho; uw < 50
If Hg is true, and since 7 is small and also o? is unknown, then the test statistics is T,

where T = x and also, T~t(n — 1)
T~t(14)

vn
x50 and) T~t(14)

ie. T = ——
5.638095238
15

Use a one-tailed test (upper tail) at 5% level
So we reject Hy if the test value of t < —1.761

= —0.4349593799
“1,761

49.73333333-50
0.43 0

t=
5.638095238
15

Since, t > —1.761, we fail to reject Hy at 5% level of significance and hence conclude

that the solicitors claim is upheld

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13. The following data give the heights in centimeters of 100 male students.

Height (cm) | Frequency

155 — 160 5
161 — 166 17
167 — 172 38
173 — 178 25
179 — 184 9
185 — 190 6

Find the expected frequencies for a normal distribution having the same mean and

standard deviation as the data given, and test the goodness of fit, using a 5% level of

significance. [13]
SOLUTION
Height (cm) Mid-points (x) Frequency (y)

154.5 — 160.5 157.5 5

160.5 — 166.5 163.5 17

166.5 — 172.5 169.5 38

172.5 — 178.5 175.5 25

178.5 — 184.5 181.5 9

184.5 — 190.5 187.5 6

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w=%=17154
6? = 7.1462641237 [from the calculator]

Ho; X~N(171.54; 7.1467)
Ho; X is not distributed this way.

Probabilities,

P(154.5 <X< 160.5) =p —-= ee)

7.146264123 7.146264123

= P(—2.384 < z < -1,545)
= 0(2.384) — 0(1.545)

= 0.9914 — 0.9388

= 0.0526 x 100 = 5.26

P(160.5 <X< 166.5) =p —- ——)

7.146264123 7.146264123
= P(-1.545 <z <—0.705)

= 0(1.545) — (0.705)

= 0.9388 — 0.7595

= 0.1793 x 100 = 17.93

P(166.5 < X <172.5) =P (

166.5-171.54 a)
7.146264123 7.146264123

= P(-0.705 <z < 0.134)
= 0(0.705) + (0.134) — 1
= 0.7595 + 0.5533 — 1

= 0.3128 x 100 = 31.28

P(172.5 <X< 178.5) =p = =)

7.146264123 Zz 7.146264123
= P(0.134 <z < 0.974)
= 6(0.974) — 6(0.134)

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= 0.8350 — 0.5533
= 0.2817 x 100 = 28.17

P(178.5 < X < 184.5) = P(

178.5-171.54 wae)
7.146264123 7.146264123

= P(0.974 <z < 1.814)
= 0(1.814) — 6(0.974)
= 0.9652 — 0.8350

= 0.1302 x 100 = 13.02

P(184.5 <X< 190.5) =p —- ee)

7ase2eti2s ~ ” ~ 7446264123.
= P(1.814 < z < 2.653)

= 0(2.653) — 0(1.814)

= 0.9960 — 0.9652

= 0.0308 x 100 = 3.08

Height (cm) Mid-points (x) Frequency (y) Expected frequencies (E)
(Observed) (0)

154.5 — 160.5 157.5 5 5.26

160.5 — 166.5 163.5 17 17.93

166.5 — 172.5 169.5 38 31.28

172.5 — 178.5 175.5 25 28.17

178.5 — 184.5 181.5 9 13.02

184.5 — 190.5 187.5 6 3.08

Degrees of freedom, (v)

There are 5 classes.

There are 3 restrictions namely;

“% >E = 100

* The mean of the distribution has been estimated from the data.

* The variance of the normal distribution has been estimated from the
data.

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«

ae

ae

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TEST 1 SOLUTION

Therefore, v = 5 — 3 = 2, hence we consider the X?(2) distribution.

Test at 5% level of significance, so rejection region is X, = 5.991.

Therefore, we reject Hy if X2,, > 5.991

cal

0 E (O—EY?
E

5 5.26 0.01285171103

17 17.93 0.04823759063

38 31.28 1.443682864

25 28.17 0.3567234647

15 16.1 0.0751552795

yO = 100

YE = 100

(0 -E)?
L—— = 193665091

Since, X2,, = 1.94 < 5.991 we fail to reject Hy, and hence conclude that data follows

anormal distribution i.e. X~N(171.54; 7.1467), at 5% level of significance.

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(— >)

FEEL FREE TO CONTACT ME FOR ANY ADJUSTMENTS, CLARIFICATIONS
AND ASSISTANCE!

schingwingwi7@gmail.com / +263784089807

Sharewell B. C

NN ,

“Concept before anything!”’, Author

Proverbs 11 vs. 2

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Complied and typed by Sharerwell B. Chingwingwi (Share)
+263784089807

schingwingwi7@qmail.com
[@2022]

NB: If you click any of the links below and it fails to direct you straight to
the respective group, then first copy the whole link of any group and,

place it in any WhatsApp/Telegram chat/group, then click again and join!

(5 WhatsApp group 1 link: https://chat.whatsapp.com/J3aylX2aHrMAdZF71Ehq5U
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