



New Member


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. 201011 to 201415).
Identify the trends in the rates of housebreaking and theft of a motor vehicle from 20111 onwards.
Rates of domestic burglary and vehicle offences for Scotland 201015
Year 
Domestic Burglary 
Theft of a motor vehicle 
201011 
0.0048 
0.0017 
201112 
0.46% 
0.13% 
201213 
0.004 
0.0011 
201314 
42 per 10,000 population 
11 per 10,0000 population 
201415 
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,



Expert


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.



New Member


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?



Expert


May 3, 2017, 09:44 AM


Not quite right  0.0001 is equivalent to 1 in 10,000, not 1 in 1,000.



New Member


May 3, 2017, 09:46 AM


Okay so it would be 0.001



Expert


May 3, 2017, 10:34 AM


Close, but they asked for 4 decimal places, not three. What should that additional digit be?



New Member


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.



New Member


May 3, 2017, 12:21 PM


Year 
Domestic burglary 
Theft of a motor vehicle 
201011 
0.0048 
0.0017 
201112 
0.0046 
0.0013 
201213 
0.0004 
0.0011 
201314 
0.0042 
0.0011 
201415 
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.



Expert


May 3, 2017, 12:38 PM


I see one error  it's in the 20122013 entry for domestic burglary.



New Member


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?



Expert


May 3, 2017, 01:02 PM


Read my previous response. You aren't focusing on the cell with the error!



New Member


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



Expert


May 3, 2017, 01:53 PM


Good!



New Member


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 201314 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 201314 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



Expert


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 20132014?



New Member


May 5, 2017, 10:08 AM


So does that mean my calculation would be 42/10,000 x 11/10,000



Expert


May 5, 2017, 11:13 AM


Correct



New Member


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



Expert


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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