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    Hi

    Can anyone help with the following:

    "The amount of fruit juice (in millimetres) in a grapefruit has a normal distribution with a mean of 90 and a variance of 100. The average amount of juice in a random sample of 25 grapefruit is A millimetres."

    Find the probability that A differs from 90 by more than 4.

    I have already calculated that N (90,4).

    Thanks
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    (Original post by flutegirl02)
    hi

    can anyone help with the following:

    "the amount of fruit juice (in millimetres) in a grapefruit has a normal distribution with a mean of 90 and a variance of 100. The average amount of juice in a random sample of 25 grapefruit is a millimetres."

    find the probability that a differs from 90 by more than 4.

    I have already calculated that n (90,4).

    Thanks
    p(|a-90| > 4) = 2p(a-90 > 4) = 2p(a > 94).
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    (Original post by Flutegirl02)
    Hi

    Can anyone help with the following:

    "The amount of fruit juice (in millimetres) in a grapefruit has a normal distribution with a mean of 90 and a variance of 100. The average amount of juice in a random sample of 25 grapefruit is A millimetres."

    Find the probability that A differs from 90 by more than 4.

    I have already calculated that N (90,4).

    Thanks
    You know that A\sim \left N( 90,100 \right ), from the information, right? Now that you have a sample, you can also assume that by \text{the Central Limit Theorem,} \ \bar{A} \approx \sim N(90,4) - it seems you've gotten that part! Now since you want to find P\left ( \left | \bar{A} - 90 \right | > 4 \right ), you can say that \bar{A} - 90 \sim N \left(0, 2^{2} \right ) by finding \mathrm{E}\left ( \bar{A} -90 \right ) and \mathrm{Var}\left ( \bar{A} - 90 \right ). This should make the rest simple from here!

    EDIT: didn't see Zacken's comment
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    (Original post by Zacken)
    p(|a-90| > 4) = 2p(a-90 > 4) = 2p(a > 94).
    Thanks Zacken
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    (Original post by aymanzayedmannan)
    You know that A\sim \left N( 90,100 \right ), from the information, right? Now that you have a sample, you can also assume that by \text{the Central Limit Theorem,} \ \bar{A} \approx \sim N(100,4) - it seems you've gotten that part! Now since you want to find P\left ( \left | \bar{A} - 90 \right | > 4 \right ), you can say that \bar{A} - 90 \sim N \left(10, 2^{2} \right ) by finding \mathrm{E}\left ( \bar{A} -90 \right ) and \mathrm{Var}\left ( \bar{A} - 90 \right ). This should make the rest simple from here!

    EDIT: didn't see Zacken's comment
    Damn... I must be forgetting my S3 already. If it's a sample, wouldn't the mean stay the same 90?
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    (Original post by Zacken)
    Damn... I must be forgetting my S3 already. If it's a sample, wouldn't the mean stay the same 90?
    made a mistake Latexing :hmmm: doing maths using it makes things me so error prone. Your method is definitely right.
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    (Original post by aymanzayedmannan)
    made a mistake Latexing :hmmm: doing maths using it makes things me so error prone. Your method is definitely right.
    Anyways, mind if I pick your brain so I understand this properly? You've got two options/methods here, right?

    You can either do P(|A -90| > 4) = 2P(\text{blah}) = \cdots like I did, or you can do:

    (A - 90) \sim N(0, 4) and then look at 2P(A > 0)?
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    (Original post by Zacken)
    Anyways, mind if I pick your brain so I understand this properly? You've got two options/methods here, right?

    You can either do P(|A -90| > 4) = 2P(\text{blah}) = \cdots like I did, or you can do:

    (A - 90) \sim N(0, 4) and then look at 2P(A > 0)?
    Suppose if they asked you to find P\left ( \left | \bar{A} -90 \right | <4 \right ). I'd find P\left ( -4 < \left (\bar{A} -90\right) < 4 \right )\] and then standardise using the distribution I've found for \bar{A} -90 previously.
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    (Original post by aymanzayedmannan)
    Suppose if they asked you to find P\left ( \left | \bar{A} -90 \right | <4 \right ). I'd find P\left ( -4 < \left (\bar{A} -90\right) < 4 \right )\] and then standardise using the distribution I've found for \bar{A} -90 previously.
    Mind writing out the standardisation for me?
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    (Original post by Zacken)
    Mind writing out the standardisation for me?
    Does it help if i let X = \bar{A} - 90 so that X\sim \left ( 0,2^{2}\right )? You only need to find P\left ( -4< X<4 \right ) then, which I'm sure will look like familiar S1 stuff!
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    (Original post by aymanzayedmannan)
    Does it help if i let X = \bar{A} - 90 so that X\sim \left ( 0,2^{2}\right )? You only need to find P\left ( -4< X<4 \right ) then, which I'm sure will look like familiar S1 stuff!
    Ah, okay, thanks - yeah that does help. Is there any advantage over that method than the standard 2P(X < 94)?
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    (Original post by Zacken)
    Ah, okay, thanks - yeah that does help. Is there any advantage over that method than the standard 2P(X < 94)?
    I've picked this up from the textbook, Sean and Gregorius. I'm not sure if it has any advantages or not :hmmmm2: Possibly in following calculations/hypothesis tests? I'm glad the OP posted this, gives me a chance to see whether or not I actually understand S3.
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    (Original post by aymanzayedmannan)
    I've picked this up from the textbook, Sean and Gregorius. I'm not sure if it has any advantages or not :hmmmm2: Possibly in following calculations/hypothesis tests? I'm glad the OP posted this, gives me a chance to see whether or not I actually understand S3.
    Same. This is cool, I always used to use my method, but yours seems like it could come in handy. Might try practising it a bit.
 
 
 
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