Help with exercise about normal distribution

I

inverse

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Hello,


I don't know much well how can I resolve these exercises.


In a particular system of units, the concentration of nitrate in a particular area of a river (near the industrial activity) follows a normal distribution with mu=8 and sigma=2.


A) Calculate the probability of which after gathering 2 samples, the levels of nitrates are superior to 10 units in someone of the samples.


B) In the same previous conditions, now with 100 samples. Calculate the probability of which in 60 or more samples it will present a concentration superior to 10 units.


Thank you very much for your help
 
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inverse said:
Hello,
not since I can calculate the following probabilities, bringing a binomial near probably?

In a particular system of units, the concentration of nitrate in a particular area of a river (near the industrial activity) follows a normal distribution with mu=8 and sigma=2.

A) To calculate the probability of which after gathering 2 samples, the levels of nitrates are superior to 10 units in someone of the samples.

B) In the same previous conditions, now with 100 samples. To calculate the probabilidad of which in 60 or more samples he will present a concentration superior to 10 units.

Thank you very much for your help
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First step is to use the normal distribution to find the probability that a single sample is 10 units or more above the mean. But if you have typed the numbers correctly, that deviation is 5*sigma and the probability is so close to zero that tables don't go that far. Please check! Once you have the correct numbers, you can find the probability p for use in the binomial distribution.

Second step is question A). You can either use the binomial formula, or you can consider the probability that neither sample is above the limit, and then subtract that from 1 to get the probability that at least one is above.

For B), since the number of samples is large, you need to use the normal approximation of the binomial distribution. That is, find mu and sigma of the binomial distribution for 100 samples, and then use the table of the normal distribution with the same mu and sigma to find the answer.

If you need more help, please show your work so we will know where you are getting stuck.
 
inverse said:
Yes, the numbers are correct
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¡Perdóname, por favor! What was wrong was the way I interpreted the numbers! It did NOT say "10 units above the mean," it said "10 units." That is, one sigma above the mean. Find p = probability that one sample from a normal distribution is greater than or equal to one standard deviation above the mean. Then proceed with the two questions.
 
inverse said:
A) Is it \(\displaystyle \binom{2}{1}p^1\cdot (1-p)^1\approx 0\)
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Once you have the correct p, that would be the probability that exactly one of the two samples is high. I believe the question wanted "at least one," in which case you should calculate probability of 0, and subtract that from 1.

P(at least 1) = 1 - P(0)
 
So the probability is at least 1.

In b) \(\displaystyle \binom{100}{1/2}p^{0.5}\cdot (1-p)^{0.5}\)

As if it approximates a normal meet that n * p ≥ 5 and do not comply with our case?

Thanks! :)
 
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DrPhil said:
¡Perdóname, por favor! What was wrong was the way I interpreted the numbers! It did NOT say "10 units above the mean," it said "10 units." That is, one sigma above the mean. Find p = probability that one sample from a normal distribution is greater than or equal to one standard deviation above the mean. Then proceed with the two questions.
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You need to go back and recalculate p.

THEN recalculate question A with the correct p

I am using upper-case "P" for the total probability, which can NEVER be greater than 1. The formula you first wrote gives probability of exactly x successes out of n trials:

\(\displaystyle \displaystyle P(x\ |\ n) = \binom{n}{x}\ p^x\ (1-p)^{(n-x)}\)

Where you said "someone of the samples" in question(A), I got the impression you might be translating from Spanish. "Una" would be "exactly one," and "alguna" would be "at least one." When dealing with "at least one," is is usually easier to calculate the probability of "none" and subtract from 1:
P(at least 1) = 1 - P(0)
 
inverse said:
As if it approximates a normal meet that n * p ≥ 5 and do not comply with our case?

Thanks! :)
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When you get p corrected, you will find that n*p ≥ 5 is satisfied for question B), with n=100.
 
But how I can calculate the P value ? if mean=8 and standard deviation=2, n=2.
 
inverse said:
But how I can calculate the p (lower-case p) value ? if mean=8 and standard deviation=2, n=2.
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You are starting with a normal distribution with mu=8 and sigma=2, but n does not enter this part of the calculation. \(\displaystyle p\) is the probability that a sample from this normal distribution is ≥ 10 units. What is the z-score for 10 if mu=8 and sigma=2? That is, how many standard deviations above the mean? Use a table of the normal distribution to find what fraction lies higher than this z-score.

Only after you know the probability for a single event can you apply the binomial distribution.

inverse said:
Is it correct? a) p(x>10)=0.2670
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No - and I can't see where that number came from.
 
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