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STANDART DEVIATION

Forums › Ask ACCA Tutor Forums › Ask the Tutor ACCA MA – FIA FMA › STANDART DEVIATION

  • This topic has 1 reply, 1 voice, and was last updated 1 year ago by John Moffat.
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  • Author
    Posts
  • February 28, 2021 at 10:37 am #612092
    suleymanabuzerli
    Member
    • Topics: 84
    • Replies: 32
    • ☆☆

    The weights of component X are normally distributed. The mean weight is 5,200kg and the standard
    deviation is 430kg.
    What is the probability of a component X weighing more than 6,000kg?
    ? 0.0314
    ? 0.2343
    ? 0.4686
    ? 0.9686

    i can’t understand this question and its explanation
    Explanation is so:

    Using z = ??
    x –
    z =
    6,000 – 5,200
    430
    z = 1.86
    z = 1.86 corresponds to an area of 0.4686. However, we are interested in the shaded area =
    0.5 – 0.4686 = 0.0314.
    If you selected 0.2343, you divided the probability obtained (0.4686) by 2 instead of subtracting
    it from 0.5.
    If you selected 0.4686, you forgot to subtract 0.4686 from 0.5.
    If you selected 0.9686, you added 0.4686 to 0.5 instead of subtracting it.

    z = 1.86 corresponds to an area of 0.4686 –i can’t understand this sentence

    1.86 corresponds to an area of 0.46 ?? but why what is logic of this?

    February 28, 2021 at 2:50 pm #612130
    John Moffat
    Keymaster
    • Topics: 57
    • Replies: 51561
    • ☆☆☆☆☆

    0.4686 comes from the tables provided in the exam and means that 0.4686 (or 48.86%) of the area under the curve lies been 5,200 and 6,000 kg. This in turn means that the is a probability of 0.4686 that a component weight between 5,200 and 6,000 kg.

    I explain how to use the tables and the logic behind it all in my free lectures on the normal distribution.

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