Maths Statistics Paper

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Question1: Frequency Distribution. Histograms, Cumulative Frequency and Ogives.

1)

a) 799
b) 1000
c) 900+9992 = 949.5
d) Lower Class Boundary=1100-0.5 =1099.5
Upper Class Boundary=1199.5+0.5=1199.5
e) 100
f) 76
g) 62400×100=15.5
h) 600-699, the fourth class has the largest frequency of 76. i)14+46+58400 ×100=29.5%
j) 48+22+6400 ×100=19.0%
k) 4002=200 200 lie in the fifth class interval 700-799 ∴The median lies in this interval. =l+n2-cff×h
=700+200-19468×99
=708.735
l)
L.Limit| U.Limit| Frequency| midpoint|  fx|
300| 399| 14| 349.5| 4893|
400| 499| 46| 449.5| 20677|
500| 599| 58| 549.5| 31871|
600| 699| 76| 649.5| 49362|
700| 799| 68| 749.5| 50966|
800| 899| 62| 849.5| 52669|
900| 999| 48| 949.5| 45576|
1000| 1099| 22| 1049.5| 23089|
1100| 1199| 6| 1149.5| 6897|
| |  ∑f 400| | ∑fx 286000|
Mean=∑fx ∑f =286000400=715
m)
Lifetime| Frequency| C. Frequency|
300-399| 14| 14|
400-499| 46| 60|
500-599| 58| 118|
600-699| 76| 194|
700-799| 68| 262|
800-899| 62| 324|
900-999| 48| 372|
1000-1099| 22| 394|
1100-1199| 6| 400|

n)
Lifetime| Frequency| C. Frequency| % C. Frequency|
300-399| 14| 14| (14 ÷ 400) x 100= 3.5 %|
400-499| 46| 60| (60 ÷ 400) x 100= 15 %|
500-599| 58| 118| (118 ÷ 400) x 100= 29.5 %|
600-699| 76| 194| (194 ÷ 400) x 100= 48.5 %|
700-799| 68| 262| (262 ÷ 400) x 100= 65.5 %|
800-899| 62| 324| (324 ÷ 400) x 100= 81 %|
900-999| 48| 372| (372 ÷ 400) x 100= 93 %|
1000-1099| 22| 394| (394 ÷ 400) x 100= 98.5 %|
1100-1199| 6| 400| (400 ÷ 400) x 100= 100 %|

o)

p)

2a&b)

Class | Frequency|
0.725-0.729| 9|
0.730-0.734| 17|
0.735-0.739| 23|
0.740-0.744| 8|
0.745-0.749| 3|

2c)

Question 2 Measures of Central Tendency

P55 is located at the 33rd ordered Term = 61

D8 is the equal to P80, which is located at the 48th Term= 79.2

Question 3 Normal Distribution
1)
Mean=625+25=650
Variance=152+32=234
Standard Deviation= 234=15.2971
×~N(650,234)
z=X-μσ=630-65015.2971=-1.3074
PX>630=1-P(X≤630)
PZ>-1.3=P(Z≤1.3)
from the normal distribution table: PZ≤1.3=0.904

2)
Mean=625 ×4=2500
Variance=225 ×4=900
Standard Deviation= 900=30
×~N(2500,900)
z=X-μσ=2450-250030=-1.667
PX>2450=1-P(X≤2450)
PZ>-1.667=P(Z≤1.667)
from the normal distribution table: PZ≤1.667=0.952
0.324]
c)
×~N(465,100)
P(22-25 years old|

Question 4 Probability Concepts
a) P(BAC of 0.08% or greater|Between 22 and 25 years old) P(A∩B)P(B)
PA∩B=0.278 ×0.114=0.039612
PB=11.4%=0.114
∴0.0396120.114=0.278=27.8%

b) P(BAC of 0.08% or greater)
=0.141×0.127+0.114×0.278+0.238×0.268+0.195×0.228+0.198×0.143+0.114×0.05 =0.191857
P(BAC of 0.08% or greater)= 19.19%

c) P(Between 22 and 25 years old|BAC of 0.08% or greater) P(B∩A)P(A)
PB∩A=0.114 ×0.278=0.039612
PA=0.191857
∴0.0396120.191857=0.20647=20.65%

d) Results to part A suggest a driver at fault who has a BAC of 0.08% or greater, has a 27.8% likelihood of being between the age of 22-25 years old. Results to part B tell us in total 19.19% of drivers from the sample taken (from all age groups) had a BAC of 0.08% or greater. Finally, results to part C suggest a driver who is at fault between the ages of 22-25 years old has a 20.65% chance of having a BAC of 0.08% or greater. Question 5 Sampling and Estimate Theories

Lower Class Limit| Upper Class Limit| Frequency| Mid Point| fx| fx2| 140| 144| 1| 144-1402=142| 142×1=142| 20164|
145| 149| 2| 149-1452=147| 147×2=294| 43218|
150| 154| 5| 154-1502=152| 152×5=760| 115520|
155| 159| 10| 159-1552=157| 157×10=1570| 246490|
160| 164| 12| 164-1602=162| 162×12=1944| 314928|
165| 169| 20| 169-1652=167| 167×20=3340| 557780|
170| 174| 24| 174-1702=172| 172×24=4128| 710016|
175| 179| 15|...
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