pauyiu 2018-12-21 10:15 PM

ppresent 2018-12-22 11:10 AM

1.將左邊變哂做base 2 先:

2^(2y)=e^x

2^(6y)=2e^(2x)=> 2^(6y-1)=e^(2x) => 2^(6y-1)=(e^x)^2

即得2^(6y-1) = (2^(2y))^2

即2^(6y-1) = 2^(4y)

即6y-1=4y

求得y之後便可ln哂兩邊得x

2. For X is a binomial variable B(n,p),

E(X)

=np

Var(X)=np(1-p)

Hence solve a simultaneous equation with

np=13.572 and

np(1-p) = 8.8489

For the mode of X you may refer to wiki:

[url]https://wikimedia.org/api/rest_v1/media/math/render/svg/74519d31faaf11d1b8e90572aeaa12fae140584e[/url]

[[i] 本帖最後由 ppresent 於 2018-12-22 11:13 AM 編輯 [/i]]

2^(2y)=e^x

2^(6y)=2e^(2x)=> 2^(6y-1)=e^(2x) => 2^(6y-1)=(e^x)^2

即得2^(6y-1) = (2^(2y))^2

即2^(6y-1) = 2^(4y)

即6y-1=4y

求得y之後便可ln哂兩邊得x

2. For X is a binomial variable B(n,p),

E(X)

=np

Var(X)=np(1-p)

Hence solve a simultaneous equation with

np=13.572 and

np(1-p) = 8.8489

For the mode of X you may refer to wiki:

[url]https://wikimedia.org/api/rest_v1/media/math/render/svg/74519d31faaf11d1b8e90572aeaa12fae140584e[/url]

[[i] 本帖最後由 ppresent 於 2018-12-22 11:13 AM 編輯 [/i]]

pauyiu 2018-12-22 07:51 PM

第2題可唔可以詳細解

ppresent 2018-12-22 10:50 PM

2.np=13.572 and

np(1-p) = 8.8489

(1-p)=8.8489/13.572

=0.65199676

p=1-0.65199676

=0.3480(corr to 4 d.p.)

n=13.572/0.348003

=38.9996

=39(corr to the nearest integer) (because n must be an integer or else X~B(n,p) is not defined)

For the mode of X~(39,0.3480), according to that of wiki (as I dun know much about it either),

because (39+1)0.3480=13.92, which is not an integer, hence the mode would be the floor(largest integer smaller than) of 13.92, which is 13.

[[i] 本帖最後由 ppresent 於 2018-12-22 10:51 PM 編輯 [/i]]

np(1-p) = 8.8489

(1-p)=8.8489/13.572

=0.65199676

p=1-0.65199676

=0.3480(corr to 4 d.p.)

n=13.572/0.348003

=38.9996

=39(corr to the nearest integer) (because n must be an integer or else X~B(n,p) is not defined)

For the mode of X~(39,0.3480), according to that of wiki (as I dun know much about it either),

because (39+1)0.3480=13.92, which is not an integer, hence the mode would be the floor(largest integer smaller than) of 13.92, which is 13.

[[i] 本帖最後由 ppresent 於 2018-12-22 10:51 PM 編輯 [/i]]

pauyiu 2018-12-23 10:53 PM

教我計埋 B(I) (II) 同埋第五題 唔該晒

ppresent 2018-12-24 07:24 PM

I think I could provide some directions for you.

4(b)(i)Assume that n nails are picked. Let the number of rusty nails in that n picked nails be X.

X~B(n,p) where p = 0.0788

To be 97% sure that there is at least 1 rusty nail (interpreted in a way that if n nails are picked at random for many many times , under 97% of the cases there is at least 1 rusty nail),it means

P(X>=1)=0.97, i.e.

1-P(X<1)=0.97

1-P(X=0)=0.97

So what is the probability P(X=0) when X~B(n,p) ? You could find that and plug it in the formula.

4(b)(ii) Similar approach: let X be the number of rusty nails in the 50 nails picked.

It is to find the probability P(X<=3)

5.(a)(i)

P(X<174.5 or X >200.5) = P(X<174.5)+P(X>200.5) as they are 2 exclusive events

Consider Z=(X-[b]μ[/b])/δ, Z~N(0,1), So you consider changing P(X<174.5) to P(Z<...),

similar for P(X>200.5)

(ii)The 72th percentile of Z~N(0,1) is 0.5828415. You could use that to find the 72th percentile of X

(b) Let X be the length of metal rods, X~N([b]μ,[/b]δ)

P(X>2261)=0.8468, i.e. P(X<2261)=0.1532

P(X>2453)=0.1182

Replace X by Z=(X-[b]μ[/b])/δ and check the standard normal table to find the value of Z with the respective probabilities, then you can find mu and sigma.

4(b)(i)Assume that n nails are picked. Let the number of rusty nails in that n picked nails be X.

X~B(n,p) where p = 0.0788

To be 97% sure that there is at least 1 rusty nail (interpreted in a way that if n nails are picked at random for many many times , under 97% of the cases there is at least 1 rusty nail),it means

P(X>=1)=0.97, i.e.

1-P(X<1)=0.97

1-P(X=0)=0.97

So what is the probability P(X=0) when X~B(n,p) ? You could find that and plug it in the formula.

4(b)(ii) Similar approach: let X be the number of rusty nails in the 50 nails picked.

It is to find the probability P(X<=3)

5.(a)(i)

P(X<174.5 or X >200.5) = P(X<174.5)+P(X>200.5) as they are 2 exclusive events

Consider Z=(X-[b]μ[/b])/δ, Z~N(0,1), So you consider changing P(X<174.5) to P(Z<...),

similar for P(X>200.5)

(ii)The 72th percentile of Z~N(0,1) is 0.5828415. You could use that to find the 72th percentile of X

(b) Let X be the length of metal rods, X~N([b]μ,[/b]δ)

P(X>2261)=0.8468, i.e. P(X<2261)=0.1532

P(X>2453)=0.1182

Replace X by Z=(X-[b]μ[/b])/δ and check the standard normal table to find the value of Z with the respective probabilities, then you can find mu and sigma.

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