5/19/2018 Chapter 3 - Representing and Summary of Data Test Solutions

Chapter 3 Representation and summary ofdata Test Solutions

1.) (a)x= 12075; x2= 15 499 685

15

12075=x = 805 B1

cao

sd =2

80515

15499685 = 620.71491 M1

& correct method

3 s.f. 621 A1

3

(NB Using n 1 gives 642.50125) (643)

(!) 99" 169" 299" 350" 475" 485" 550" 650" 689" 830" M1

999" 1015" 1050" 2100" 2315

Attempt to order

Q2= 650 A1

cao 650

IQR= Q3 Q1= 1015 350 = 665

Attempt at Q3 Q1cao 665

4

(#) Q3$ 1.5(Q3 Q1) = 1015 $ 1.5 % 665 = 2012.5 M1

Use of given outier formua

Q1 1.5(Q3 Q1) = 350 1.5 % 665 & 0 M1

!vidence "oth ends considered

2100 and 2315 a'e *+ie's A13

(d)

, e ! s i * e

- . ( /

0 5 0 0 1 0 0 0 1 5 0 0 2 0 0 0 2 5 0 0

!/+*s B1

sae s#a+e

!* +a!e++ed

5/19/2018 Chapter 3 - Representing and Summary of Data Test Solutions

,e!si*e B1

-/ B/+* B1

B* *+ie's B1

4

NB ' s/" 'ig* !and ise' d'an * 2012.5 is a##e/*a!+e.

(e) Median e!si*e edian s/

,e!si*e nega*ive se; s/ a//' se*'i#a+

:gn'ing *+ie's

anges a//'ia*e+ e

5/19/2018 Chapter 3 - Representing and Summary of Data Test Solutions

a*ien*s% (420) and (390) a'e *+ie's. B1f*B1f*

4

M1 f' a #+ea' a**e/* sing *ei'

5/19/2018 Chapter 3 - Representing and Summary of Data Test Solutions

4. (a) Q2=2

1616+= 16; Q1= 15; Q3= 16.5; : = 1.5

M1A1; B1; B1;

B1 (5)

(") 1.5 % : = 1.5 % 1.5 = 2.25 M1 A1

Q1 1.5 % : = 12.75 n *+ie's !e+ Q1 A1Q3$ 1.5 % : = 18.75 25 is an *+ie' A1B/+*" +a!e+ s#a+e M1

14" 15" 16" 16.5" 18.75 (18) A1@*+ie' A1 (7)

(c) x =20

322= 16.1 M1 A1 (2)