Statistics and Probability Homework: Math Assignment Solution, 2019

This document presents a detailed solution to a mathematics assignment centered on statistics and probability. The assignment covers a wide range of topics, including the application of the normal distribution to calculate probabilities, binomial distribution problems, and hypothesis testing scenarios. It includes calculations involving z-scores, probabilities of events occurring within specific ranges, and the use of statistical formulas to solve real-world problems. The solution provides step-by-step explanations for each problem, demonstrating the application of statistical concepts to various scenarios such as quality control, financial analysis, and population studies. The document also explores the concepts of probability, including calculating probabilities for defective items in a shipment, and evaluating the likelihood of events occurring in a given sample. The document also includes problems involving the calculation of probabilities for events, such as the probability of a batch being rejected based on the number of defective items found in a sample. The solution further addresses the use of statistical concepts to analyze and interpret data, and includes the use of statistical concepts to analyze and interpret data, and make conclusions about the population. This assignment solution is ideal for students seeking help with their statistics and probability homework.

Mathematics Assignment
Student Name:
Instructor Name:
Course Number:
10th July 2019

1a)
z= x−μ
σ
When x=1.72, μ=1.5 , σ =0.1
z= 1.72−1.5
0.1 =2.2
P(X>1.72)=P(Z>2.2)=1-(0.4861+0.5)=0.0139
P(X>1.72) =0.0139
b)
z= x−μ
σ
When x=1.35, μ=1.5 , σ =0.1
z= 1.35−1.5
0 .1 =−1.5
P(Z<-1.5)=0.5-0.4332=0.0668
P(X<1.35)= P(Z<-1.5)=0.0668
c)
z= x−μ
σ
When x=1.45, μ=1.5 , σ =0.1
z= 1.45−1.5
0.1 =−0.5
P(Z=-0.5)=1-0.6915=0.3185
When x=1.62, μ=1.5 , σ =0.1
z= 1.62−1.5
0.1 =1.2
P(Z=1.2)=0.8849
P(1.45≤X≤ 1.62¿=¿P(-0.5≤Z≤1.2)=0.8849-0.3185=0.5664
P(1.45 ≤X≤ 1.62¿=¿P(-0.5≤Z≤1.2)=0.1915+0.3849=0.5764
2.
P(defective)=0.15

P(0 or 1 defective)= P(0 defective)+ P(1 defective)
10C0(0.15)0(0.85)10 +10C1(0.15)1(0.85)9
=0.1969+0.3474=0.5443
P(accepting shipment)=0.5443
3.a)
mean= 400+3800
2 =¿2100
mean =$2100
b)
standard deviation= b−a
√12 = 3800−400
√12 = 3400
√12 =981.5
standard deviation=$ 981.5
c)
3800-400=3400
b-a=2000-400=1600
P(less than $2000)= 1600
3400 =0.4706
P(less than 2000) ¿ 0.4706
d)
3800-3000=800
P(more than $3000)= 800
3400 =0.2353
P(more than $3000)¿ 0.2353
e)
P(exactly $2500)¿ 1−¿ P(more than $2500)+ P(less than $ 2500) ¿
P(exactly $2500)¿ 1−¿ 3800−2500
3400 + 2500−400
3400
P(exactly $2500) ¿ 1− ( 1300
3400 + 2100
3400 ) =0

4. )
a) z= x−μ
σ
When x=300, μ=250 , σ =2500
z= 300−250
2500 =0.02
P(x>300)=P(Z>0.02)=0.5-0.008=0.492
b)
When x=190, μ=250 , σ =2500
z= 190−250
2500 =−0.024
When x=270, μ=250 , σ =2500
z= 270−250
2500 =0.008
P(190<x<270)=P(-0.024¿Z<0.008)=(0.008+0.004)=0.012
c)
When x=260, μ=250 , σ =2500
z= 260−250
2500 =0.004
P(x ≤260)=P(Z ≤0.02)=
d) 90
100 ×250 hours=225 hours
e) 0.05×250hours=12.5 hours
5. a)i
P(between 15 and 20 defective items)= P(15 defective ) or P(16 defective)or P(17 defective)or P(18
defective)or P(19 defective)or P(20 defective)

P(between 15 and 20 defective items)= 100C15(0.10)15(0.90)85 +100C16(0.10)16(0.90)84+
+100C17(0.10)17(0.90)83+100C18(0.10)18(0.90)82+100C19(0.10)19(0.90)81+100C20(0.10)20(0.90)80
=0.0327+0.0193+0.0106+0.0054+0.0026+0.0012=0.0718
P(between 15 and 20 defective items)= 0.0718
ii)
P(at most 5 defective)= P(0 defective ) or P(1 defective)or P(2 defective)or
P(3 defective)or P(4 defective)or P(5 defective)
=100C0(0.10)0(0.90)100 +100C1(0.10)1(0.90)99+
+100C2(0.10)2(0.90)98+100C3(0.10)3(0.90)97+100C4(0.10)4(0.90)96+100C5(0.10)5(0.90)95
=0.000027+0.000295+0.00162+0.005892+0.015875+0.033866=0.057575
= P(at most 5 defective)= 0.057575
iii)
P( at least 20 defective)=1- P(20 defective)
P(20 defective)= 100C20(0.10)20(0.90)80=0.001171
P( at least 20 defective)=1- P(20 defective)=1-0.001171=0.998829
P( at least 20 defective) =0.998829
b)
Middle 80% will fall from 40% t 60%
40% 0f 100=40
60% 0f 100=60
Range if from 40 to 60 items
6) a)
z= x−μ
σ
When x=800, μ=550 , σ =150
z= 800−550
150 =1.667

P(X>800)=P(Z>1.667)=0.5-0.4525=0.0475
b)
z= x−μ
σ
When x=450, μ=550 , σ =150
z= 450−550
150 =−0.667
When x=550, μ=550 , σ =150
z= 550−550
150 =0
P(450¿ x <¿550)=P(-0.667 ¿ z <¿0)=0.2486+0=0.2486
c)
z= x−μ
σ
When x=?, μ=550 , σ =150
When p=0.05,Z=0.13
0.13= x−550
150
x-150=0.13(150)
x=150+19.5=169.5
Minimum daily balance= $ 169.5
7a)
P(all 7 serving at least 24 hours)=(0.7)7=0.082354
b)
P(at least 3 serving at least 24 hours)=P(3 serving at least 24 hours)or P(4 serving at least 24 hours)or
P(5 serving at least 24 hours)or P(6 serving at least 24 hours)or P(7 serving at least 24 hours)or
=(0.7)3+(0.7)4+(0.7)5+(0.7)6++(0.7)7=0.951173
P(at least 3 serving at least 24 hours)= 0.951173
8a)

a=0.5, b=10
b)
mean= a+b
2 =0.5+10
2 =5.25 minutes
Standard deviation= b−a
√12 = 10−0.5
√12 = 9.5
√12 =2.74 minutes
c)
10-5=5
10-0.5=9.5
P(more than 5 minutes)= 5
9.5 = 10
19
d)
Middle 50% will fall from 25% t 75%
10-0.5=9.5
25% 0f 9.5=2.375
75% 0f 9.5=7.125
2.375+0.5=2.875
7.125+0.5=7.625
Range if from 2.875 to 7.626 minutes
9a)
z= x−μ
σ
When x=0.2, μ=0.15 , σ =0.02
z= 0.2−0.15
0.02 =2.5
P(X ≥ 0.2 ¿=¿P(Z≥ 2.5 ¿=0.5-0.4938=0.0062
b)
P(1 out of 4 obese)=0.25×0.0062=0.00155
c) 50
1000 =0.005
0.005×0.0062=0.000031

d)
The fact that probability of being obese (0.0062) before survey is greater than the probability of an
individual picked from 10000 people having more than 20% or more having body fat(0.003),we can
conclude that the army is successful in reducing percentage of obese men below the percentage in the
general population
10.a)
P( foreign made)= 2
5 =0.4
P( no more than 8 foreign)= P(0 foreign) + P(1 foreign) + P(2 foreign) + P(3 foreign) + P(4 foreign) + P(5
foreign) + P(6 foreign) + P(7 foreign) + P(8 foreign)
=15C0(0.4)0(0.6)15+15C1(0.4)1(0.6)14+15C2(0.4)2(0.6)13+15C3(0.4)3(0.6)12+15C4(0.4)4(0.6)11+15C5(0.4)5(0.6)10+
15C6(0.4)6(0.6)9+15C7(0.4)7(0.6)8+15C8(0.4)8(0.6)7
=0.00047+0.004702+0.021942+0.0063388+0.126776+0.185938+0.206598+0.177084+0.118056
=0.847905
=P( no more than 8 foreign)= 0.847905
b)
P( more owners of foreign than American cars)=1-0.4=0.6
c)
P( exactly 10 foreign cars)= 15C10(0.4)10(0.6)5=0.024486
P( exactly 10 foreign cars)= 0.024486
11. a)
z= x−μ
σ
When x=45, μ=36 , σ = √36=6
z= 45−36
6 =1.5
P(X¿ 45 ¿=P(Z¿ 1.5 ¿=0.5-0.4332=0.0668
b)
When x=30, μ=36 , σ =6
z= 30−36
6 =−1

When x=39, μ=36 , σ =6
z= 39−36
6 =0.5
P(30<X<39)=P(-1<Z<0.5)=0.3413+0.1915=0.5328
c)
When x=30, μ=36 , σ =6
z= 30−36
6 =−1
P(30<X)= P(-1<Z)=0.5-0.3413=0.1587
12.
P(this many or more defective)= 2
100× 55
2500= 110
250000 = 11
25000
13a)
P(free beer)= 30
100 =0.3
P(free cola )= 65
100 =0.65
P( exactly 5 beer coupon)= 12C5(0.3)5(0.65)7=0.094346
P( exactly 5 beer coupon)=0.094346
b)
P( at least 3 beer coupon)=1- P(less than3 beer coupon)
P(less than3 beer coupon)= P(0 beer coupon)+ P(1 beer coupon)+ P(2 beer coupon)
=12C0(0.3)0(0.65)12 +12C1(0.3)1(0.65)11 +12C2(0.3)2(0.65)10
=0.005688+0.031503+0.079969=0.11716
P( at least 3 beer coupon)=1- P(less than3 beer coupon)=1-0.11716=0.88284
P( at least 3 beer coupon)= 0.88284
c)

P( at least 8 cola beer coupon)= P(8 cola beer coupon)+ P(9 cola beer coupon)+ P(10 cola beer coupon)+
P(11 cola beer coupon)+ P(12 cola beer coupon)
=12C8(0.65)8(0.3)4+ 12C9(0.65)9(0.3)3+12C10(0.65)10(0.3)2+12C11(0.65)11(0.3)1+12C12(0.65)12(0.3)0
0.127761+0.123029+0.079969+0.031503+0.005688=0.36795
P( at least 8 cola beer coupon)= 0.36795
d)
P( between 5 and 7beer coupon)= P(5 beer coupon)+ P(6 beer coupon)+ P(7 beer coupon)
=12C5(0.65)7(0.3)5+ 12C6(0.65)6(0.3)6+12C7(0.65)5(0.3)7
=0.094346+0.050802+0.020097=0.165245
P( between 5 and 7beer coupon)= 0.165245
14 a)
P(100 or more recommending)= 50
150 × 60
100 = 1
5
b)P(more than 85)= 65
150 × 60
100 = 13
50
P(less than 95)= 94
150 × 60
100 = 47
125
P(more than 85 but less than 95)= 47
125 −13
50 = 29
250
15
a)
P(100 or more voters in favour)= 100
200 =1
2 =0.5
P( endorsing if 45 percent favour it)= P( endorsing and 45 percent favour it)=0.5×0.45=0.225
P( endorsing if 45 percent favour it)=0.225
b)
P( not endorsing if 52 percent don’t favour it)=P(not endorsing and 52 percent don’t favour
it)=0.5×0.52=0.26
P( not endorsing if 52 percent don’t favour it)=0.26
16.a)

P(shutting down in the next four weeks)=(0.2)5=0.00032
b)
P(shutting down in the next four weeks)= )=(0.2)6=0.000064
c)
P( A batch rejected)=P(2 or more defective from sample and batch in bad condition)
= 9
10 × 15
100 = 135
1000 =0.135
P( A batch rejected)= 0.135
d)
P( A batch accepted)= P(less than2 defective from sample and batch in good condition)
1
10 × 85
100 = 85
1000 =0.085
17 a)
z= x−μ
σ
Top 10%=1-10%=90%=0.9
When P=0.9,Z=4.28
When x=?, μ=11000 , σ =2500
4.28= X−11000
2500
X-11000=2500(4.28)
X=11000+2500(4.28)=21700
To be in top 10% you must earn from $21700
b)
Middle 50% starts from 25% to 75%
When p=25%=0.25,Z=0.67
z= x−μ
σ

0.67= x−11000
2500
x-11000=0.67(2500)
x=1100+0.67(2500)=12675
Minimum income= $12 675
18.a)
P(admitting no more than 2)= P(admitting 0)or P(admitting 1)or P(admitting 2) 12C8(0.65)8(0.3)4+
15C0(0.2)0(0.8)15+15C1(0.2)1(0.8)14+15C2(0.2)2(0.8)13
=0.035184+0.131941+0.230897=0.398022
P(admitting no more than 2)= 0.398022
b)
P(admitting exactly 1)= 15C1(0.2)1(0.8)14=0.131941
P(admitting exactly 1)= 0.131941
19.i)
P(at least 6 women)=P(6 women) or P(7 women) or P(8 women) or P(9 women) or P(10 women)
= 10C6(0.3)6(0.7)4+10C7(0.3)7(0.7)3+10C8(0.3)8(0.7)2+10C9(0.3)9(0.7)1+10C10(0.3)10(0.7)0
=0.036757+0.009002+0.001447+0.000138+0.000006=0.04735
P(at least 6 women)= 0.04735
ii)
P(no more than 3 women)= p(0 women) or P(1 women) or P(2 women) or P(3 women)
=10C0(0.3)0(0.7)10+10C1(0.3)1(0.7)9+10C2(0.3)2(0.7)8+10C3(0.3)3(0.7)7
=0.028248+0.121061+0.233474+0.266828=
iii)
P( at least 2 but more than 8)=P( 9 women) or P( 10 women)
=10C9(0.3)9(0.7)1+10C10(0.3)10(0.7)0
=0.000138+0.000006=
vi)
P(exactly 5 men)= 10C5(0.3)5(0.7)5=0.102919