# Maths Statistics Paper

**Topics:**Nontotient, Normal distribution, Hebrew numerals

**Pages:**5 (905 words)

**Published:**April 1, 2013

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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