# normal distribution

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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 Ïƒ . 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.

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.

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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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