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**jasminM01****Guest**

Linda took a multiple choice test she did pretty well, but there were 4 questions where she had to guess. Each question had 5 choices for answers. Make up a probability distribution for the number of correct guesses. What is the probability that she got at least one right? what is the probability that she didn't get them all right? what is the mean and standard deviation for the number of correct guesses?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,656

I do not have to make a distribution there is already one. It is called the Binomial distribution.

P(1 or more right ) = 369 / 625

P(none right ) = 1 - (369 / 625)

P(all right ) = 1 / 625

P(she did not get them all right) = 1 - ( 1 / 625 )

The mean of the binomial distribution is np and the sd is √ (npq)

Can you fill in the rest?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**jasminM01****Guest**

your a life saver!!!!!!!

THANK YOU

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,656

Hi;

Welcome to the forum. Lifeguards save lives, firemen and paramedics do too. There is a bit more to that question, did you get it?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**jasminM01****Guest**

yea i think i can probably figure it out after what you have answered. its my take home midterm and my teacher wasn't very helpful in explaining it.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,656

yea i think i can probably figure

I like the use of the word probably in a probability question. Very good!

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**jasminM01****Guest**

Haha i didn't notice that until you pointed it out!

**jasminM01****Guest**

jasminM01 wrote:

Haha i didn't notice that until you pointed it out!

can you actually help me with the rest of it cause i can't figure it out!

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,656

Hi;

The mean of the binomial distribution is np and the sd is √ (npq)

n is the number of trials

p is the probability of success.

q = 1 - p

Are you okay with this?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**jasminM01****Guest**

jasminM01 wrote:

jasminM01 wrote:Haha i didn't notice that until you pointed it out!

can you actually help me with the rest of it cause i can't figure it out!

so would n=20 because thats how many choices you have

and p= 4

q=1-4=3

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,656

Hi;

p = 1 / 5 because that is the probability of her getting a question right.

q = 1 - 1 / 5 = 4 / 5

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

jasminM01 wrote:

jasminM01 wrote:jasminM01 wrote:Haha i didn't notice that until you pointed it out!

can you actually help me with the rest of it cause i can't figure it out!

so would n=20 because thats how many choices you have

and p= 4

q=1-4=3

You have 4 questions, each of which she has a 0.2 probability of answering correctly...

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