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    About time someone posted a stats question up here.
    In the book Heinemann Modular Mathematics for Edexcel AS and A Level
    Statistics 3
    Exercise 1A Question 8

    A battery under normal usage has a mean life of 160 hours, with a standard deviation of 30 hours. Batteries are bought in packs of four. A tourist has a camera that takes one battery and he hopes that the pack of four will last for at least 700 hours. Find the probability of this being the case.

    So for one battery it is normally distributed with X~N(160,30²)
    right?
    So for a pack of batteries of 4, then it must be Y~N(640,120²)?
    or am I wrong?

    It should be here where it has caused my error. Can someone help me? Thank you
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    A battery under normal usage has a mean life of 160 hours, with a standard deviation of 30 hours. Batteries are bought in packs of four. A tourist has a camera that takes one battery and he hopes that the pack of four will last for at least 700 hours. Find the probability of this being the case.

    So for one battery it is normally distributed with X~N(160,30²) right?
    Right
    So for a pack of batteries of 4, then it must be Y~N(640,120²)?
    or am I wrong?
    Wrong.

    Let X be the DRV 'the life of a battery', then:

    X ~ N(160, 30²)
    700/4 = 175
    P(X > 175) = 1 - Φ((175-160)/30)
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    But that still doesn't explain why the answer in the back is:
    0.1587
    1 - Φ((175-160)/30)
    = 1- Φ(0.5)
    = 1- 0.6915
    = 0.3085
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    Don't you have to times the variance by 42 ?

    EDIT: You did that. Still thinking...
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    (Original post by samd)
    Don't you have to times the variance by 42 ?
    Yeah you do
    But that would give you the distribution of
    X~N(640,120²)
    or am I wrong?
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    (Original post by kenobi124)
    Yeah you do
    But that would give you the distribution of
    X~N(640,120²)
    or am I wrong?
    Even if that distribution was right, the SD should not be greater than 60.
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    (Original post by kenobi124)
    Yeah you do
    But that would give you the distribution of
    X~N(640,120²)
    or am I wrong?
    Use:

    X ~ N(640, 60²)
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    (Original post by Dekota)
    Use:

    X ~ N(640, 60²)
    May I ask why?
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    (Original post by kenobi124)
    May I ask why?
    I can't follow this either. On pg 2 of S3 it says "Var(aX) = a2Var(X)".
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    (Original post by kenobi124)
    May I ask why?
    If you have the distribution: X ~ N(160, 30²)

    then for n number of X, you have X ~ N(160n, (30√n)²)

    so in this case n = 4, and so X ~ N(640, (60)²)

    SOrry if it doesn't make sense, but yay I finised stats!
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    (Original post by kenobi124)
    About time someone posted a stats question up here.
    In the book Heinemann Modular Mathematics for Edexcel AS and A Level
    Statistics 3
    Exercise 1A Question 8

    A battery under normal usage has a mean life of 160 hours, with a standard deviation of 30 hours. Batteries are bought in packs of four. A tourist has a camera that takes one battery and he hopes that the pack of four will last for at least 700 hours. Find the probability of this being the case.

    So for one battery it is normally distributed with X~N(160,30²)
    right?
    So for a pack of batteries of 4, then it must be Y~N(640,120²)?
    or am I wrong?

    It should be here where it has caused my error. Can someone help me? Thank you
    There are 4 batteries in the pack and one battery usually has a mean life of 160 hours.
    Let B be the random variable of 4 batteries life.
    E(B) = 160 + 160 + 160 + 160
    = 160 . 4
    = 640
    Var(B) = 30^2 + 30^2 +30^2 +30^2
    = 30^2 . 4
    = 3600
    So, standard deviation(B)= sqt root 3600
    = 60
    P(x>_B) = P(Z>_ [700-640]/60)
    = P(Z>_1)
    = 1 - 0.8413
    = 0.1587

    This is not multiple random variables so you could not use Var(ax+b) = a^2Var(x).
 
 
 
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