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    nargis123's Avatar
    nargis123 Posts: 28, Reputation: 1
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    #1

    May 2, 2017, 08:42 AM
    Finance
    Transform the figures in the table into one comparable measure of risk (probability of the offence expressed as a decimal, rounded to four decimal places) keeping the rows in the same year order (i.e. 2010-11 to 2014-15).
    Identify the trends in the rates of housebreaking and theft of a motor vehicle from 201-11 onwards.

    Rates of domestic burglary and vehicle offences for Scotland 2010-15
    Year Domestic Burglary Theft of a motor vehicle
    2010-11 0.0048 0.0017
    2011-12 0.46% 0.13%
    2012-13 0.004 0.0011
    2013-14 42 per 10,000 population 11 per 10,0000 population
    2014-15 1 in 256 1 in a thousand

    Please can someone help me as to what needs to be done. I am not asking for the answers but I am confused as to what I exactly need to do. Can someone please provide me with an example with the figures above so I know what is being asked for,
    ebaines's Avatar
    ebaines Posts: 12,130, Reputation: 1307
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    #2

    May 2, 2017, 01:08 PM
    The first row of data presents numbers as decimals to four significant digits - for example 0.0048. That's the form that they want all the numbers in. Looking at row 2 there is an entry of 0.46% -- if you convert that to a decimal you get 0.0046, right? Another example: the last row includes 1 in a thousand, which is 1/1000; can you write that out as a decimal? Post back with yourr attempts at this and we'll check it for you.
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    #3

    May 3, 2017, 07:53 AM
    Hi thank you for responding to my post. So the 1/1000 would that be 0.0001 if its to four decimal places?
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    ebaines Posts: 12,130, Reputation: 1307
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    #4

    May 3, 2017, 09:44 AM
    Not quite right - 0.0001 is equivalent to 1 in 10,000, not 1 in 1,000.
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    #5

    May 3, 2017, 09:46 AM
    Okay so it would be 0.001
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    #6

    May 3, 2017, 10:34 AM
    Close, but they asked for 4 decimal places, not three. What should that additional digit be?
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    #7

    May 3, 2017, 11:03 AM
    Oh I forgot to put the 0 at the end so it would be 0.0010.

    I will have a go at the rest of the figures and will post up here and will appreciate if you could look over for me.
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    #8

    May 3, 2017, 12:21 PM
    Year Domestic burglary Theft of a motor vehicle
    2010-11 0.0048 0.0017
    2011-12 0.0046 0.0013
    2012-13 0.0004 0.0011
    2013-14 0.0042 0.0011
    2014-15 0.0039 0.0010

    I have completed the table (attempted), please can you have a look at it and let me know if this is correct.

    Thank you once again.
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    ebaines Posts: 12,130, Reputation: 1307
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    #9

    May 3, 2017, 12:38 PM
    I see one error - it's in the 2012-2013 entry for domestic burglary.
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    #10

    May 3, 2017, 12:55 PM
    There was a typo on the original table posted it should say '11 per 10,000'. I accidentally added another zero. However my answer was still correct I'm assuming.
    The '42 per 10,000' I calculated as 42/10000 which gave me 0.0042? Did I do my calculations wrong?
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    ebaines Posts: 12,130, Reputation: 1307
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    #11

    May 3, 2017, 01:02 PM
    Read my previous response. You aren't focusing on the cell with the error!
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    #12

    May 3, 2017, 01:12 PM
    Sorry my mistake I thought you had said the cell below. I see the error it should be 0.0040
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    #13

    May 3, 2017, 01:53 PM
    Good!
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    #14

    May 4, 2017, 09:32 AM
    With the second question on my coursework they have said:

    'Explain what is meant if two events are described as statistically independent. What was the chance of suffering from housebreaking and theft of a motor vehicle in 2013-14 assuming they were unrelated risks? Answer can be expressed as a decimal or a percentage to at least one decimal figure'

    The first half I have put down:

    Two or more events are said to be statistically independent if the occurrence of one event does not affect the probability of the other event(s). For example, the probability of student A passing an examination is not dependent on whether student B passes or not. These are statistically independent events.

    I am not sure what the second part of the question is asking. So this is the probability of both offences happening in 2013-14 if I'm correct?
    So if I look at the domestic burglary entry from the original table would it be as simple as 42/10,000 x 100 = 0.42% (if showing as a percentage).

    Please could you give you input. Thank you
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    #15

    May 4, 2017, 11:21 AM
    If the probability of event A happening is give as p, and the probability of event B happening is q, and they are independent events, then the probability of both A and B happening is p x q. For example: if the probability of being born male is 1/2, and the probability of being born with red hair is 1/20, and if we assume those events are independent, then the probability of being born a male with red hair is 1/2 x 1/20 = 1/40. Now, how would you apply this to determining the probability of experiencing both a domestic burglary and a motor vehicle theft in 2013-2014?
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    #16

    May 5, 2017, 10:08 AM
    So does that mean my calculation would be 42/10,000 x 11/10,000
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    #17

    May 5, 2017, 11:13 AM
    Correct
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    #18

    May 5, 2017, 11:29 AM
    462/100,000,000 = 231/50,000,000 is this correct?

    So as a percentage would it be 0.000462% so to 1 significant figure would be 0.50% - I am not sure if this half is correct
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    #19

    May 5, 2017, 11:41 AM
    Well, 0.5% is over 1000 times greater than 0.000462%. It's also greater than either of the two probabilities that you're working with, so it should be clear that it can't be correct. If you round 0.000462% to one significant digit you get 0.0005%. Do you see how that works?

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