Statistical Inference

[deelaw007]

Please answer the following questions. From q.41 onwards

 

[Eugene99]

do you have the answers?

 

[deelaw007]

for odd numbers only
like 41, 43 like that

 

[Eugene99]

is difficlt then, because I work backwards
but post the answers, I will try

 

[Eugene99]

Question 42:
a) How many of these new cardholders would you expect to default? What is the standard deviation?
µ = 12 * (0.07) = 0.84
σ = sqrt[(12) * (0.07) * (0.93)] = 0.8839

b) What is the likelihood that NONE of the cardholders will default?
0.4186, found by (12!/(0! * 12!)) *(0 .07)^0 * (0.93)^12

c) What is the likelihood that AT LEAST ONE will default?
0.4186, found by 1 - .4186 = .5814


Question 43:


Question 44:
a) X ~ Bin(20, 0.01). Probability of X being 0 or 1 is (0.99)^20 + 20*0.01*(0.99)^19.

b) X ~ Bin(20, 0.02). etc...

c) X ~ Bin(20, 0.05). etc...

 

[deelaw007]

More answers?

 

[deelaw007]

Well these are correct.

 

[Eugene99]

Question 45:
Choosing 2 people from the 6 that liked the new flavor and 2 people from the 4 that did not. This can be done in
C(6,2) * C(4,2) = 15*6 = 90 ways.

Now since there are C(10,4) = 210 ways we can pick 4 of the 10 people in the group without restrictions, the desired probability is
90/210 = 3/7.

Question 46:
Binomial Distribution with n= 10, p = 0.45, q = 0.55
The expansion is (p+q)^10
a
Mean = np = 4.5
SD = sqrt(npq) = sqrt(10 x 0.55 x 0.45) = 1.573

b
Find term containing p^4 in the binomial expansion
This is 10C4 x 0.45^4 x 0.55^6
= 210 x 0.0410 x 0.0277
= 0.2383

c
P(x < 5) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)
= 10c0*(0.45)^0*(1-0.45)^10 + 10c1*(0.45)^1*(1-0.45)^9 + 10c2*(0.45)^2*(1-0.45)^8
+ 10c3*(0.45)^3*(1-0.45)^7 + 10c4*(0.45)^4*(1-0.45)^6
= 0.5044

 

[deelaw007]

Question 45:
Choosing 2 people from the 6 that liked the new flavor and 2 people from the 4 that did not. This can be done in
C(6,2) * C(4,2) = 15*6 = 90 ways.

Now since there are C(10,4) = 210 ways we can pick 4 of the 10 people in the group without restrictions, the desired probability is
90/210 = 3/7.

Question 46:
Binomial Distribution with n= 10, p = 0.45, q = 0.55
The expansion is (p+q)^10
a
Mean = np = 4.5
SD = sqrt(npq) = sqrt(10 x 0.55 x 0.45) = 1.573

b
Find term containing p^4 in the binomial expansion
This is 10C4 x 0.45^4 x 0.55^6
= 210 x 0.0410 x 0.0277
= 0.2383

c
P(x < 5) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)
= 10c0*(0.45)^0*(1-0.45)^10 + 10c1*(0.45)^1*(1-0.45)^9 + 10c2*(0.45)^2*(1-0.45)^8
+ 10c3*(0.45)^3*(1-0.45)^7 + 10c4*(0.45)^4*(1-0.45)^6
= 0.5044

Thank you bro
50, 51 and 52?

 

[deelaw007]

Question 45:
Choosing 2 people from the 6 that liked the new flavor and 2 people from the 4 that did not. This can be done in
C(6,2) * C(4,2) = 15*6 = 90 ways.

Now since there are C(10,4) = 210 ways we can pick 4 of the 10 people in the group without restrictions, the desired probability is
90/210 = 3/7.

Question 46:
Binomial Distribution with n= 10, p = 0.45, q = 0.55
The expansion is (p+q)^10
a
Mean = np = 4.5
SD = sqrt(npq) = sqrt(10 x 0.55 x 0.45) = 1.573

b
Find term containing p^4 in the binomial expansion
This is 10C4 x 0.45^4 x 0.55^6
= 210 x 0.0410 x 0.0277
= 0.2383

c
P(x < 5) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)
= 10c0*(0.45)^0*(1-0.45)^10 + 10c1*(0.45)^1*(1-0.45)^9 + 10c2*(0.45)^2*(1-0.45)^8
+ 10c3*(0.45)^3*(1-0.45)^7 + 10c4*(0.45)^4*(1-0.45)^6
= 0.5044

Dont do Q49

 

[Eugene99]

Question 50:
a)
n= 200 ; p = 0.015 (defective) ; q = 0.895 (not defective)
1. None of the antennas is defective.
P(200 not defective) = 0.895^200 = 2.315.10^-10

b)
Three or more of the antennas are defective
P(3<= x <=200)
= 1 - P(0<= x <= 2)
= 1 - binomcdf(200,0.015,2)
= 0.5785

Question 51:
a)
e(-4)=0.01832 That is the probability of 0.
b)
probability of 4= e (-4) 4^4/4! = 0.2930 (expected value)
c)
probability of 4 or fewer is 0.629 from table
d)
probability of 4 or more are waiting is 0.371 (from ≥4 + 0.2930 (4 exactly)=0.664


Question 52:
a)
P(1,2) = [e^(-2)*2^1]/1! = 2e^(-2) = 0.2707
b)
1-poissoncdf(2,4)=0.05265...
c)
poissonpdf(2,0) = 0.13534...

=======================
Cheers!

 

[deelaw007]

Few more questions :3

 

[deelaw007]

Cont

 

[deelaw007]

Question 50:
a)
n= 200 ; p = 0.015 (defective) ; q = 0.895 (not defective)
1. None of the antennas is defective.
P(200 not defective) = 0.895^200 = 2.315.10^-10

b)
Three or more of the antennas are defective
P(3<= x <=200)
= 1 - P(0<= x <= 2)
= 1 - binomcdf(200,0.015,2)
= 0.5785

Question 51:
a)
e(-4)=0.01832 That is the probability of 0.
b)
probability of 4= e (-4) 4^4/4! = 0.2930 (expected value)
c)
probability of 4 or fewer is 0.629 from table
d)
probability of 4 or more are waiting is 0.371 (from ≥4 + 0.2930 (4 exactly)=0.664


Question 52:
a)
P(1,2) = [e^(-2)*2^1]/1! = 2e^(-2) = 0.2707
b)
1-poissoncdf(2,4)=0.05265...
c)
poissonpdf(2,0) = 0.13534...

=======================
Cheers!

Thanks man :')

 

[Eugene99]

Cont

Question 37:
binom dist with n = 12, p = 0.45, q = 0.55

E[x] = np = 12*0.45 = 5.4 <------ ans to part 1

08 C(12,08) * 0.45^08 * 0.55^4 0.07616 -----> 0.0762 (ans to part 2)
09 C(12,09) * 0.45^09 * 0.55^3 0.02770
10 C(12,10) * 0.45^10 * 0.55^2 0.00680
11 C(12,11) * 0.45^11 * 0.55^1 0.00101
12 C(12,12) * 0.45^12 * 0.55^0 0.00007

for part 3, add up and round to 4 dp

Question38:
Binomial Problem with n = 12 and p = 0.30

P(x = 2) = 12C2(0.3)^2*(0.7)^10 = binompdf(12,0.3,2) = 0.1678 = 16.78%

So the probabiity of that happenin is close to 17%.

Question 39:
a
expected value = np = 15*0.4 = 6

b
P(x = 10) = 15C10*0.4^10*0.6^5 = binompdf(15,0.4,10) = 0.0245

c
P(10<= x <= 15) = 1 - binomcdf(15,0.4,9) = 0.0338

d
P(11<= x <=15) = 1 - binomcdf(15,0.4,10) = 0.0093

==============================

Please excuse me from Question 35 and 36.

 

[deelaw007]

Question 37:
binom dist with n = 12, p = 0.45, q = 0.55

E[x] = np = 12*0.45 = 5.4 <------ ans to part 1

08 C(12,08) * 0.45^08 * 0.55^4 0.07616 -----> 0.0762 (ans to part 2)
09 C(12,09) * 0.45^09 * 0.55^3 0.02770
10 C(12,10) * 0.45^10 * 0.55^2 0.00680
11 C(12,11) * 0.45^11 * 0.55^1 0.00101
12 C(12,12) * 0.45^12 * 0.55^0 0.00007

for part 3, add up and round to 4 dp

Question38:
Binomial Problem with n = 12 and p = 0.30

P(x = 2) = 12C2(0.3)^2*(0.7)^10 = binompdf(12,0.3,2) = 0.1678 = 16.78%

So the probabiity of that happenin is close to 17%.

Question 39:
a
expected value = np = 15*0.4 = 6

b
P(x = 10) = 15C10*0.4^10*0.6^5 = binompdf(15,0.4,10) = 0.0245

c
P(10<= x <= 15) = 1 - binomcdf(15,0.4,9) = 0.0338

d
P(11<= x <=15) = 1 - binomcdf(15,0.4,10) = 0.0093

==============================

Please excuse me from Question 35 and 36.

More Questions coming up

cause quiz is on wednesday xD

 

[deelaw007]

From 17-38

 

[Eugene99]

From 17-38

Q17:
μ=50, σ=4
a)
Z(44) = (44-50)/4 = -1.5
Z(55) = (55-50)/4 = 1.25
From the table, 1.5 is 0.9332 so 1-0.9332 = 0.0668
1.25 is 0.8944
44<Z<55 : 0.8944-0.0668
=0.8275

b)
Z(55) as calculated above:
1- z(55)
= 1-0.8944
=0.1056

c)
Z(52) = (52-50)/4 = 0.5 from the table : 0.5 is 0.6915
Z(55) again : 0.8944
52<Z<55 = 0.8944 - 0.6915
= 0.2029
_______________________________________________________________
doing the rest, lettme know if the answers are wrong

 

[Eugene99]

Q18:
μ=80, σ=14
a) Z(75) = (75-80)/14 = -0.36 from the table: 0.36: 0.6406
1-0.6406
= 0.36
Z(90) = (90-80)/14 = 0.714 from the table: 0.714: 0.7611
so, 75<Z<90 is 0.7611-0.36 = 0.40

b) It's exactly like the one in previous question. Since we already found Z(75) which is 0.36
Less than 75 will be 1 - 0.36 = 0.64

c) Z(55) = (55-80) / 14 = -1.78 from the table 1.78 is 0.9625
1-0.9625
=0.03
Z(70) = (70-80)/14 = -0.71 from the table 0.71 is 0.7661
1-0.7661
=0.23
Now, from 55 to 70: 0.23-0.03
=0.20

 

[Eugene99]

Question 19:
μ=2708, σ=650
a)
Z(3000) = (3000-2708)/650 = 0.4492 from the table: 0.4492: 0.6736
1-0.6736
= 0.326
For percentage, just multiply the probability by 100
0.326 will be 32.6%

b)
We already found Z(3000) in the previous part, so,
Z(3500) = (3500-2708)/650 = 1.21
1.21 from the table: 0.8869
between 3000 and 3500: 0.8869 - 0.6736
= 0.2133
= 21.33%

c) Z(2500) = (2500-2708)/ 650 = -0.32
0.32 from the table is 0.6255
since it was negative value we subtract from 1 :
1-0.6255
= 0.3745
We already found Z(3500) in previous part, which was 0.8869
Between 3500 and 2500:
0.8869 - 0.3745
= 0.514
or say 51.4%