# question on aircraft bombing target using normal/gaussian distribution sampling

• Jul 3rd 2010, 08:00 PM
gkraju
question on aircraft bombing target using normal/gaussian distribution sampling
assume the dispersion of 3000 bombs launched from an aircraft showed a normally distributed miss distance profile with standard deviation of 10 meters. if 80 bobmbs are dropped on a target with no systematic errors at launch , what is probability that 50 bombs are within 10 meters of target
• Jul 4th 2010, 03:28 AM
mr fantastic
Quote:

Originally Posted by gkraju
assume the dispersion of 3000 bombs launched from an aircraft showed a normally distributed miss distance profile with standard deviation of 10 meters. Mr F says: And what's the mean? Are we meant to assume it's zero?

if 80 bobmbs are dropped on a target with no systematic errors at launch , what is probability that 50 bombs are within 10 meters of target

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• Jul 4th 2010, 03:32 AM
CaptainBlack
Quote:

Originally Posted by gkraju
assume the dispersion of 3000 bombs launched from an aircraft showed a normally distributed miss distance profile with standard deviation of 10 meters. if 80 bobmbs are dropped on a target with no systematic errors at launch , what is probability that 50 bombs are within 10 meters of target

This is about as elementary a problem on the normal distribution as you can get, so what have you done and what difficulties are you having (you might also go back to you teacher and tell them to word their problems more carefully, the first sentence if taken literally is nonsense and they should know better).

CB
• Jul 4th 2010, 04:23 AM
gkraju
In the question, mean was not stated directly i think we have to find the answer assuming mean to be zero.
i have no clue how to go ahead with it, i tried this way, assuming mean to be zero,

calculating z = (X- mean)/ std deviation, i.e. z = 10-0/10 = 1.
then i calculate area under the normal curve between -1 to 1.
now i get the probability of bomb hitting target with in + or - 10 metres, this is for population of 3000 bombs.

i am struck here, how to make this probability which i got for 3000 bombs to be applicable for getting
50bombs withi in 10 metres out of 80 bombs ?
• Jul 4th 2010, 06:20 AM
CaptainBlack
Quote:

Originally Posted by gkraju
In the question, mean was not stated directly i think we have to find the answer assuming mean to be zero.
i have no clue how to go ahead with it, i tried this way, assuming mean to be zero,

calculating z = (X- mean)/ std deviation, i.e. z = 10-0/10 = 1.
then i calculate area under the normal curve between -1 to 1.
now i get the probability of bomb hitting target with in + or - 10 metres, this is for population of 3000 bombs.

i am struck here, how to make this probability which i got for 3000 bombs to be applicable for getting
50bombs withi in 10 metres out of 80 bombs ?

If your teacher were doing their job properly the question would read:

Assume the dispersion of bombs launched from an aircraft is normally distributed about the aim point with standard deviation of 10 meters. if 80 bombs are dropped on a target with no systematic errors at launch , what is probability that 50 bombs are within 10 meters of target. (I will assume we have a one dimensional error here in reallity we would be dealing with a 2-D normal error distribution)

The probability that a single bomb is within 1 sd of the target is ~0.6827

Now that exactly 50 of 80 are within 1sd is given by the binomial distribution b(50;80,0.6827), which you evaluate either with a binomial calculator or using the normal approximation.

Now if the wording of the question is sloppy and really means 50 or more you have to sum b(k;80,0.6827) for k=50 to 80, or again make use of the normal approximation.

CB
• Jul 4th 2010, 08:00 AM
gkraju
" Now that exactly 50 of 80 are within 1sd is given by the binomial distribution b(50;80,0.6827), which you evaluate either with a binomial calculator or using the normal approximation. "

can u please clarify what is binomial distribution (50 : 80, 0.6827 ) ?
• Jul 4th 2010, 08:44 AM
SpringFan25
You may have seen a different notation: $X \sim Bi(80, 0.6827)$

find $P(X=50)$

You should have been taught how to do this before being set the question
• Jul 4th 2010, 08:46 AM
CaptainBlack
Quote:

Originally Posted by gkraju
" Now that exactly 50 of 80 are within 1sd is given by the binomial distribution b(50;80,0.6827), which you evaluate either with a binomial calculator or using the normal approximation. "

can u please clarify what is binomial distribution (50 : 80, 0.6827 ) ?

See the Wikipedia page. Note they use f(k,N,p) for the probability mass function of the binomial distribution rather than b(k,N,p) as I do.

CB