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I am working with Nov 2014 paper 2, and I had a problem to understand and  solve a) (ii), can some one explain it to me? 

 

 

The weights of fish in a lake are normally distributed with a mean of 760g and
 
standard deviation σ . It is known that 78.87% of the fish have weights between
 
705g and 815 g.
 
(a) (i) Write down the probability that a fish weighs more than 760g.
 
(ii) Find the probability that a fish weighs less than 815 g.
 
 
Thanks in advance! 
Edited by Maha

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705 and 815 are both 55 from the mean. Meaning that from 705 to 760 the probability is (78.87/2) and same for 760 to 815. 

 

a) Probability of weighing more than 760 is 0.5, or 50% since the area on either side of the mean is 50%.

 

ii) Since the area from negative infinity to 760 is 50% and 760 to 815 is (78.87/2) the total area up to 815 is 89.435. Hence the probability that the fish weighs less than 815 is 0.894.

Edited by dorianb
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Yeah, it took some time for me to understand the question as well.

 

(a)

i. Well we know that if 760 g is the mean weight of the fish, that means that half of them have a weight under 760 g and half of them above 760 g. Therefore the P (X > 760) = 0.5

 

ii. For this one, you may want to draw out the normal distribution curve. So it'll look like this (it's REALLY rough, forgive me for the bad drawing)

 

http://tinyurl.com/n949us7 

 

So you know that P (705<X<815) = 0.7887

 

Recognize that the portions on either side of the mean are symmetrical. So now we can subtract 0.7887 from 1 to find the remaining values for the two unknown sections (the sections in blue). To find ONE section simply divide this value by 2. 

 

(1 - 0.7887) / 2

= 0.10565 (the section in green)

 

And now add this to the 0.7887 to get the probability of fish that weigh less than 815 g

 

Therefore, P (X < 815) = approx. 0.894 (significant digits)

 

Edit: Sorry! I was working on it and I guess dorianb already posted

Edited by IbTrojan
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I posted worked solutions to the November exam here:

 

http://bit.ly/ib-sl-maths-nov-2014

 

Hope this helps.

Thank you very much , it helped me a lot. 

 

705 and 815 are both 55 from the mean. Meaning that from 705 to 760 the probability is (78.87/2) and same for 760 to 815. 

 

a) Probability of weighing more than 760 is 0.5, or 50% since the area on either side of the mean is 50%.

 

ii) Since the area from negative infinity to 760 is 50% and 760 to 815 is (78.87/2) the total area up to 815 is 89.435. Hence the probability that the fish weighs less than 815 is 0.894.

Thanks for the help :)

 

Yeah, it took some time for me to understand the question as well.

 

(a)

i. Well we know that if 760 g is the mean weight of the fish, that means that half of them have a weight under 760 g and half of them above 760 g. Therefore the P (X > 760) = 0.5

 

ii. For this one, you may want to draw out the normal distribution curve. So it'll look like this (it's REALLY rough, forgive me for the bad drawing)

 

http://tinyurl.com/n949us7 

 

So you know that P (705<X<815) = 0.7887

 

Recognize that the portions on either side of the mean are symmetrical. So now we can subtract 0.7887 from 1 to find the remaining values for the two unknown sections (the sections in blue). To find ONE section simply divide this value by 2. 

 

(1 - 0.7887) / 2

= 0.10565 (the section in green)

 

And now add this to the 0.7887 to get the probability of fish that weigh less than 815 g

 

Therefore, P (X < 815) = approx. 0.894 (significant digits)

 

Edit: Sorry! I was working on it and I guess dorianb already posted

Thank you as well :) 

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