Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Stats Quesition . Normal Distribution.

  • 19-04-2013 11:14am
    #1
    Registered Users, Registered Users 2 Posts: 5,068 ✭✭✭


    Hi Folks ,


    Stuck on this question in revision.


    Weight of packet of crisps are normally distributed with a mean of 35g and standard dev. of 1.1 g .

    i) find the prob. that a packet at random weighs more than 36g.

    Fine with that one. Did as follows-
    X= RV amount of crisps in packet. / x N ( 35 , 1.1 )

    and then that was P(x > 36 ) = 1 - P( x ≤ 36 ) . Got that as being 0.8289 . I think thats correct.



    Part 2 im stuck on .


    ii) Find The probability that the combined weight of 10 packets is less than 355g.

    Can't find any examples in my notes for this one.



    ANy help much appreciated . Thanks. :)


Comments

  • Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    1) You might want to look at your tables again. You want 1-[LATEX]\Phi(\frac{36-35}{1.1})[/LATEX]. Intuitively, the probablility of something being *greater* than the mean cannot be greater than 0.5...

    2) Sums of normally distributed variables are normally distributed with a mean of the sum of the means of the components and a *variance* (NB not standard deviation) of the sum of the variances of the components (so the sd is the a square root of the sum of the variances).
    Bearing that in mind, then, how is the weight of 10 packets of crisps distributed?


  • Registered Users, Registered Users 2 Posts: 5,068 ✭✭✭Iancar29


    Yakuza wrote: »
    1) You might want to look at your tables again. You want 1-[LATEX]\Phi(\frac{36-35}{1.1})[/LATEX]. Intuitively, the probablility of something being *greater* than the mean cannot be greater than 0.5...

    2) Sums of normally distributed variables are normally distributed with a mean of the sum of the means of the components and a *variance* (NB not standard deviation) of the sum of the variances of the components (so the sd is the a square root of the sum of the variances).
    Bearing that in mind, then, how is the weight of 10 packets of crisps distributed?

    Ah cheers thanks i forgot the 1- (.8289) part.

    Ah so then it would be 10 times the original mean ( 350) . Then for the SD i First square the original SD . ( 1.21 ) ... x 10 = 12.1 an get the sq. root of that then = 3.48?

    I'll try with those figures now so.. cheers


Advertisement