Turn on thread page Beta
    • Thread Starter
    Offline

    4
    ReputationRep:
    The random variable X has a B(40. 0.3) distribution. The mean of a random sample of n observations of X is denoted by x̄. Find(a) P(x̄≥13) when n is 49,(b) the smallest value of n for which P(x̄≥13)<0.001.
    I've done part a and for b got upto (13)-(1 divided by 2n)-(2) all divided by (square root 8.4 divided by square root n)=3.090
    How would i obtain a value for n?
    • Thread Starter
    Offline

    4
    ReputationRep:
    for more clarity here's a picture
    Attached Images
     
    Offline

    18
    ReputationRep:
    isnt  \displaystyle \bar{x} \sim (40, \sqrt{\frac{0.3}{n}}^2) and  \displaystyle P(\bar{x} &gt; X) = P(Z &gt; \frac{X - 40}{\sqrt\frac{{0.3}}{n}})

    (just checking I have this right before I go any further)
    • Thread Starter
    Offline

    4
    ReputationRep:
    Wouldn't the variance be 8.4 over n, since mean for binomial approximated to normal is npq, so 40*0.3*0.7=8.4
    Offline

    18
    ReputationRep:
    (Original post by gooner1010)
    Wouldn't the mean be 8.4 over n, since mean for binomial approximated to normal is npq, so 40*0.3*0.7=8.4
    hmm as I feared, I cant remember how to do this :hide: sorry

    Edit: Is this S2? a binomial to normal approximation question? ill revise it and get back to you if you want?
    • Thread Starter
    Offline

    4
    ReputationRep:
    (Original post by DylanJ42)
    hmm as I feared, I cant remember how to do this :hide: sorry

    Edit: Is this S2? a binomial to normal approximation question? ill revise it and get back to you if you want?
    Hi, yep it's a binomial approximation to normal, and help would be appreciated.
    Thanks
    Offline

    18
    ReputationRep:
    (Original post by gooner1010)
    Hi, yep it's a binomial approximation to normal, and help would be appreciated.
    Thanks
    sorry I honestly dont have a clue how to do this. Hopefully someone else can help you
    Offline

    18
    ReputationRep:
    (Original post by gooner1010)
    Hi, yep it's a binomial approximation to normal, and help would be appreciated.
    Thanks
    last try, is your answer to the first part 0.8869?

    Spoiler:
    Show

    please say yes
    • Thread Starter
    Offline

    4
    ReputationRep:
    (Original post by DylanJ42)
    last try, is your answer to the first part 0.8869?

    Spoiler:
    Show


    please say yes

    Sorry. it's 0.0084
    Offline

    18
    ReputationRep:
    (Original post by gooner1010)
    Sorry. it's 0.0084
    ive got no idea then, sorry :laugh:

    iirc SeanFM knows his stats, maybe he can salvage this
    Offline

    14
    ReputationRep:
    (Original post by gooner1010)
    The random variable X has a B(40. 0.3) distribution. The mean of a random sample of n observations of X is denoted by x̄. Find(a) P(x̄≥13) when n is 49,(b) the smallest value of n for which P(x̄≥13)<0.001.
    I've done part a and for b got upto (13)-(1 divided by 2n)-(2) all divided by (square root 8.4 divided by square root n)=3.090
    How would i obtain a value for n?
    So this is a normal approximation to the binomial type question. If you have a random variable X distributed as B(n,p) then it is approximated by a variable distributed as N(np, np(1-p)) (where the second parameter for the normal is given as the variance rather that standard deviation).

    So here, X is approximated by N(40 * 0.3, 40 * 0.3 * 0.7) = N(12, 8.4)

    If X is distributed as N(\mu, \sigma^2), then X-bar is distributed as N(\mu, \sigma^2/n) where n is the number of observations. So here we have the mean distributed as N(12, 0.1714286). So part (a) follows by looking at tables or using software.

    If n is unknown, then you need to work in reverse, try a few values of n and see what values you get for the probability - after a while you should be able to pinpoint the n you want.
    • Thread Starter
    Offline

    4
    ReputationRep:
    (Original post by Gregorius)
    So this is a normal approximation to the binomial type question. If you have a random variable X distributed as B(n,p) then it is approximated by a variable distributed as N(np, np(1-p)) (where the second parameter for the normal is given as the variance rather that standard deviation).

    So here, X is approximated by N(40 * 0.3, 40 * 0.3 * 0.7) = N(12, 8.4)

    If X is distributed as N(\mu, \sigma^2), then X-bar is distributed as N(\mu, \sigma^2/n) where n is the number of observations. So here we have the mean distributed as N(12, 0.1714286). So part (a) follows by looking at tables or using software.

    If n is unknown, then you need to work in reverse, try a few values of n and see what values you get for the probability - after a while you should be able to pinpoint the n you want.
    The book has the answer 82, however I seem to be getting 81
    Offline

    14
    ReputationRep:
    (Original post by gooner1010)
    The book has the answer 82, however I seem to be getting 81
    I agree with your value of 81. I'm using the following R code

    N <- 81
    mm <- 12
    ss <- (40 * 0.3 * 0.7)/N
    1 - pnorm(13, mm, sqrt(ss))
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 31, 2017
Poll
“Yanny” or “Laurel”
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.