The standard deviation if the number of defective rivets in a randomly selected seam? [closed]

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

An aircraft seam requires 10 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of each other, each with the same probability. Suppose further that 20% of all seams end up having to be reworked. What is the standard deviation of the number of defective rivets in a randomly selected seam?
The standard deviation of a binomial distribution with n=10 and p=0.2 is equal to 1.26 but the correct answer is 0.46.

probability

share|cite|improve this question

asked Jul 26 at 20:35

Roy Rizk

887

closed as off-topic by heropup, John Ma, max_zorn, Taroccoesbrocco, amWhy Jul 30 at 11:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, John Ma, max_zorn, Taroccoesbrocco, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
0
down vote

favorite

An aircraft seam requires 10 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of each other, each with the same probability. Suppose further that 20% of all seams end up having to be reworked. What is the standard deviation of the number of defective rivets in a randomly selected seam?
The standard deviation of a binomial distribution with n=10 and p=0.2 is equal to 1.26 but the correct answer is 0.46.

probability

share|cite|improve this question

asked Jul 26 at 20:35

Roy Rizk

887

closed as off-topic by heropup, John Ma, max_zorn, Taroccoesbrocco, amWhy Jul 30 at 11:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, John Ma, max_zorn, Taroccoesbrocco, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

An aircraft seam requires 10 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of each other, each with the same probability. Suppose further that 20% of all seams end up having to be reworked. What is the standard deviation of the number of defective rivets in a randomly selected seam?
The standard deviation of a binomial distribution with n=10 and p=0.2 is equal to 1.26 but the correct answer is 0.46.

probability

share|cite|improve this question

asked Jul 26 at 20:35

Roy Rizk

887

An aircraft seam requires 10 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of each other, each with the same probability. Suppose further that 20% of all seams end up having to be reworked. What is the standard deviation of the number of defective rivets in a randomly selected seam?
The standard deviation of a binomial distribution with n=10 and p=0.2 is equal to 1.26 but the correct answer is 0.46.

probability

share|cite|improve this question

asked Jul 26 at 20:35

Roy Rizk

887

share|cite|improve this question

share|cite|improve this question

asked Jul 26 at 20:35

Roy Rizk

887

asked Jul 26 at 20:35

Roy Rizk

887

asked Jul 26 at 20:35

Roy Rizk

887

887

closed as off-topic by heropup, John Ma, max_zorn, Taroccoesbrocco, amWhy Jul 30 at 11:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, John Ma, max_zorn, Taroccoesbrocco, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by heropup, John Ma, max_zorn, Taroccoesbrocco, amWhy Jul 30 at 11:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, John Ma, max_zorn, Taroccoesbrocco, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

The seam will have to be reworked if any of these rivets is defective.

Suppose rivets are defective independently of each other, each with
the same probability. Suppose further that 20% of all seams end up
having to be reworked

These are the crucial information to answer the question. If at least one of the rivets are defective then the seam will have to be reworked. Let $p$ the probability that an arbitrarily rivet is defective. Let X be the r.v. for the defective rivets. Then we have

$$P(Xgeq 1)=1-P(X=0)=1-(1-p)^10=0.2$$

Solve the equation for $1-p$ and $p$ respectively.

And finally $sigma_x=sqrtncdot pcdot (1-p)$

share|cite|improve this answer

edited Jul 26 at 21:29

answered Jul 26 at 21:07

callculus

16.4k31427

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

The seam will have to be reworked if any of these rivets is defective.

Suppose rivets are defective independently of each other, each with
the same probability. Suppose further that 20% of all seams end up
having to be reworked

These are the crucial information to answer the question. If at least one of the rivets are defective then the seam will have to be reworked. Let $p$ the probability that an arbitrarily rivet is defective. Let X be the r.v. for the defective rivets. Then we have

$$P(Xgeq 1)=1-P(X=0)=1-(1-p)^10=0.2$$

Solve the equation for $1-p$ and $p$ respectively.

And finally $sigma_x=sqrtncdot pcdot (1-p)$

share|cite|improve this answer

edited Jul 26 at 21:29

answered Jul 26 at 21:07

callculus

16.4k31427

add a comment | 

up vote
1
down vote



accepted

The seam will have to be reworked if any of these rivets is defective.

Suppose rivets are defective independently of each other, each with
the same probability. Suppose further that 20% of all seams end up
having to be reworked

These are the crucial information to answer the question. If at least one of the rivets are defective then the seam will have to be reworked. Let $p$ the probability that an arbitrarily rivet is defective. Let X be the r.v. for the defective rivets. Then we have

$$P(Xgeq 1)=1-P(X=0)=1-(1-p)^10=0.2$$

Solve the equation for $1-p$ and $p$ respectively.

And finally $sigma_x=sqrtncdot pcdot (1-p)$

share|cite|improve this answer

edited Jul 26 at 21:29

answered Jul 26 at 21:07

callculus

16.4k31427

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

The seam will have to be reworked if any of these rivets is defective.

Suppose rivets are defective independently of each other, each with
the same probability. Suppose further that 20% of all seams end up
having to be reworked

These are the crucial information to answer the question. If at least one of the rivets are defective then the seam will have to be reworked. Let $p$ the probability that an arbitrarily rivet is defective. Let X be the r.v. for the defective rivets. Then we have

$$P(Xgeq 1)=1-P(X=0)=1-(1-p)^10=0.2$$

Solve the equation for $1-p$ and $p$ respectively.

And finally $sigma_x=sqrtncdot pcdot (1-p)$

share|cite|improve this answer

edited Jul 26 at 21:29

answered Jul 26 at 21:07

callculus

16.4k31427

The seam will have to be reworked if any of these rivets is defective.

Suppose rivets are defective independently of each other, each with
the same probability. Suppose further that 20% of all seams end up
having to be reworked

These are the crucial information to answer the question. If at least one of the rivets are defective then the seam will have to be reworked. Let $p$ the probability that an arbitrarily rivet is defective. Let X be the r.v. for the defective rivets. Then we have

$$P(Xgeq 1)=1-P(X=0)=1-(1-p)^10=0.2$$

Solve the equation for $1-p$ and $p$ respectively.

And finally $sigma_x=sqrtncdot pcdot (1-p)$

share|cite|improve this answer

edited Jul 26 at 21:29

answered Jul 26 at 21:07

callculus

16.4k31427

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 21:29

edited Jul 26 at 21:29

edited Jul 26 at 21:29

answered Jul 26 at 21:07

callculus

16.4k31427

answered Jul 26 at 21:07

callculus

16.4k31427

answered Jul 26 at 21:07

callculus

16.4k31427

16.4k31427

add a comment | 

add a comment |