the u and sigma of wing length(a normally distributed parameter) in a population of fruit flies are 4 and 0.2mm respectively. In a random sample of 400 fruit flies, how many individuals are expected to have wing lengths greater than 4.4mm ?
1. 20
2. 64
3. 10
4. 336

Question

the u and sigma of wing length(a normally distributed parameter) in a population of fruit flies are 4 and 0.2mm respectively.  In a random sample of 400 fruit flies, how many individuals are expected to have wing lengths greater than 4.4mm ?
1.  20
2.  64
3.  10
4.  336

abhisar · Answer

this one @vivek.pattani

anonymous · Answer

ok no problem

aaronq · Answer

u is the mean?

4.4 mm, is correspondent to $2\sigma$=0.4 mm, which is 95% 

so 5% will have wings>4.4mm

0.05*400=20

aaronq · Answer

wait i actually think its should be half of that, because the 5% takes into account both ends of the distribution. So it's 10

anonymous · Answer

thanks
but can u explain me in detail

aaronq · Answer

I have to go for now but i can explain later. In the mean time, is there anything is specific you dont understand?

anonymous · Answer

yes  statistics

aaronq · Answer

lol statistics in general?

well i can explain how to solve the problem if you'd like me to do that... but if you want a deeper understanding you need to read about it in a book or online somewhere.

anonymous · Answer

just tell me about the question how did u solve

aaronq · Answer

|dw:1400770589489:dw|

In a normal (Gaussian) distribution, $\sigma=68\%$, $2\sigma=95\%$

the mean is 4 mm, so $\pm0.2mm$ would be 1 standard deviation. meaning 68% of measured wings will be between 3.8 mm and 4.2 mm.

$\pm0.4mm$ would be two standard deviations, so 95% of measured wings will be between 3.6 mm and 4.4 mm.

The question only asks about the upper end.|dw:1400770931080:dw|

This upper end is half of the rest (from 95%), which is 2.5%.

so 2.5% of 400 individuals is 10.