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# Find the median mass of a medium size apple Watch

1. Apples are Normally distributed by mass. M~N(110, 15^2)

A medium sized apple is:

90 <= x < 120

Question is:

Find the median mass of a medium size apple

My attempt was:

(120 + 90) / 2 = 105g

Seemed reasonable.

But in fact answer is 107g. Why and how?
2. (Original post by acomber)
Apples are Normally distributed by mass. M~N(110, 15^2)

A medium sized apple is:

90 <= x <= 120

Question is:

Find the median mass of a medium size apple

My attempt was:

(120 + 90) / 2 = 105g

Seemed reasonable.

But in fact answer is 107g. Why and how?
Why would that be correct

Your working assumes an even distribution between 90 and 120 but that is not the case
3. (Original post by TenOfThem)
Why would that be correct

Your working assumes an even distribution between 90 and 120 but that is not the case
But I thought the median was the middle value?
4. (Original post by acomber)
But I thought the median was the middle value?
Yes, it is, but you do not just have 2 apples

Is the median of 1,2,2,7,7,8,8,8,9 the same as (1+9)/2
5. (Original post by acomber)
But I thought the median was the middle value?
Yes, but if you have: 12, 12, 15, 24. What's the median?
6. (Original post by TenOfThem)
Yes, it is, but you do not just have 2 apples

Is the median of 1,2,2,7,7,8,8,8,9 the same as (1+9)/2
I see...

Can you give me a hint on how to calculate then?
7. (Original post by Andy98)
Yes, but if you have: 12, 12, 15, 24. What's the median?
(12+15)/2 = 13.5.

So I have to find out roughly the middle in the distribution? How do I do that?
8. (Original post by acomber)
I see...

Can you give me a hint on how to calculate then?
I am sure there is an annoying formula somewhere
9. (Original post by acomber)
(12+15)/2 = 13.5.

So I have to find out roughly the middle in the distribution? How do I do that?
You first have to find the area between when x=120 and x=90 then the median will lie at half that area from both x=90 and x=120 use z=(x-mean)/standard deviation to calculate the areas using tables and then you will have to work backwards to find value of x when that area is the area at x=120 -half the area between x=120 and x=90 and then rearrange for x.
10. (Original post by Dalek1099)
You first have to find the area between when x=120 and x=90 then the median will lie at half that area from both x=90 and x=120 use z=(x-mean)/standard deviation to calculate the areas using tables and then you will have to work backwards to find value of x when that area is the area at x=120 -half the area between x=120 and x=90 and then rearrange for x.
OK got it.

Area up to 90 = 0.0913
Area under 90 to 120 = 0.6564

Half area between 90 to 120 is median value = 0.6564/2 = 0.3282

0.0913 + 0.3282 = 0.4195

We now just find the z value for 0.4195

phi(z) = 0.4195

tables only give +ve z so lookup for 1-0.4195=0.5805, that z = 0.2045

Therefore, z for our area = -0.2045

-0.2045 = (x - 110)/15 x = 106.93 ~ 107.

Perfect.

Thanks.

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