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Stats Quesition . Normal Distribution.

  • 19-04-2013 11:14AM
    #1
    Iancar29
    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

  • #2
    Yakuza
    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?


  • #3
    Iancar29
    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


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