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    Re: Please can you check my answers




    Is that all parts in Q1?

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    Re: Please can you check my answers

    Quote Originally Posted by zkm1223 View Post
    I have shown my working in the attachments in posts 7, 8 and 9.
    My bad.
    ***************x

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    Re: Please can you check my answers

    Quote Originally Posted by zkm1223 View Post
    Is that all parts in Q1?
    Yup. I didn't see anything wrong with it.

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    Re: Please can you check my answers

    Quote Originally Posted by zkm1223 View Post
    Q2) An electrical appliance lifetime follows a normal distribution with mean 530 hrs and SD 51 hrs. Calculate:

    a) percentage of components that fail in under 600 hrs (My answer: 8.53%)
    This isn't right. It's always good to do a common sense check: is it really likely that 8.53% of components are going to fail in 600 hours, when the mean is considerably less than 600? I think you're looking at the wrong side of the distribution; you need the percentage of cases that fall under a z score of 1.37, not the percentage above it.

    b) percentage that fail between 471 and 525 hrs (My answer: 33.72%)
    Looks right. You can check these quite easily yourself using this tool.

    c) lifetime that components can exceed with 95% confidence (My answer 613.64 hrs)
    I think this is right.

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    Re: Please can you check my answers

    This isn't right. It's always good to do a common sense check: is it really likely that 8.53% of components are going to fail in 600 hours, when the mean is considerably less than 600? I think you're looking at the wrong side of the distribution; you need the percentage of cases that fall under a z score of 1.37, not the percentage above it.
    Is the answer 91.47%?

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    Re: Please can you check my answers

    Quote Originally Posted by zkm1223 View Post
    Is the answer 91.47%?
    Use the link I posted to check.

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    Re: Please can you check my answers

    Ok, thanks. Please can you check the final question.

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    Re: Please can you check my answers

    It's definitely wrong. Are you sure you used the right numbers for the mean and sd?

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    Re: Please can you check my answers

    here is my working for Q3
    Attached Images  

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    Re: Please can you check my answers

    Well. Um. That does shed some light on the problem. When you're deciding whether or not you're looking at a one-tail or a two-tail problem it doesn't change what the mean and standard deviation are. The stuff you have labeled '2way' is just completely wrong. Completely forget anything you were thinking when you decided to go that route. Do the problem using the 1way numbers.

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    Re: Please can you check my answers

    I have kept the mean and SD the same, however I get a z value of 5.6. This value is not on my tables so what should I do?
    Attached Images  

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    Re: Please can you check my answers

    I still don't understand how to do this last question

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    Re: Please can you check my answers


    I think your first attempt is very close.

    Let X_1, X_2 be the travelling time to and from work respectively.

    Then X_1 \stackrel{d}{=} X_2 \sim \mathcal{N}(\mu = 67, \sigma^2 = 15^2)

    Moreover we assume X_1, X_2 are independent.

    Then the total working time Y \triangleq X_1 + X_2 is also
    normally distributed.

    To determine the distribution of Y, you just need to compute
    E[Y], Var[Y]

    Note that Var[Y] = Var[X_1] + Var[X_2] since they are independent,

    and thus SD[Y] = \sqrt{Var[X_1] + Var[X_2]}

    Now you want to evaluate the probability \Pr\{Y > 151 \}

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