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Normal Distribution Samples

  • Thread starter DotFourier
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  • #1

Homework Statement


Suppose that diastolic blood pressures (DBPs) for men aged 35-44 are normally distributed with a mean of 80 (mm Hg) and a standard deviation of 10. What's the probability that in a random sample of 5 subjects, 4 or more have DBPs more than 90?

Homework Equations




The Attempt at a Solution


I first thought of finding the probability that someone from the sample of 5 has DBP greater than 90. After converting to standard normal and looking at a Z table I found that it was 0.0125. I do not know where to go from here. I am confused as to what to do to be able to find the probability of 4 or more having more than 90 DBP.
 

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  • #2
LCKurtz
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Homework Statement


Suppose that diastolic blood pressures (DBPs) for men aged 35-44 are normally distributed with a mean of 80 (mm Hg) and a standard deviation of 10. What's the probability that in a random sample of 5 subjects, 4 or more have DBPs more than 90?

Homework Equations




The Attempt at a Solution


I first thought of finding the probability that someone from the sample of 5 has DBP greater than 90. After converting to standard normal and looking at a Z table I found that it was 0.0125. I do not know where to go from here. I am confused as to what to do to be able to find the probability of 4 or more having more than 90 DBP.
Start by finding the probability that 1 person has DBP > 90. I would say ##.0125## looks way too small. How did you get that?
 
  • #3
RUber
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Once you have the right probability for one sample, then you can use a combinatorial approach.
Let p(x) be the probability that any one sample has DBP>90 and p(~x) be the probability of not that.
p(~x)=1-p(x).
Choose 5 people at random, there are ##\begin{pmatrix} 5\\4\end{pmatrix} ## ways to get 4 people with high blood pressure, and the number of ways to have 4 or more is slightly greater than that.
You could use a probability tree, but that would take a long time.

You could also enumerate all the options and find the probability of each case:
X1 = x, x, x, x, x (5 people have high BP) , p(X1) = ?
x, x, x, x, ~x (4 people have high BP)
.
.
.
~x, x, x, x, x (4 people have high BP)
The total probability is the sum of the independent cases.

However, using the fact that you can know how many cases of each type there might be, and the relative similarity of the probability profile for each case, you can simplify this down to a quick and easy function of p(x).
 
  • #4
Ray Vickson
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Homework Statement


Suppose that diastolic blood pressures (DBPs) for men aged 35-44 are normally distributed with a mean of 80 (mm Hg) and a standard deviation of 10. What's the probability that in a random sample of 5 subjects, 4 or more have DBPs more than 90?

Homework Equations




The Attempt at a Solution


I first thought of finding the probability that someone from the sample of 5 has DBP greater than 90. After converting to standard normal and looking at a Z table I found that it was 0.0125. I do not know where to go from here. I am confused as to what to do to be able to find the probability of 4 or more having more than 90 DBP.
This is like a coin-tossing problem, but with a biased coin having ##P(\text{heads}) = p##. Here, ##p## is the probability that an individual has DBP > 90; you claimed that ##p = 0.0125## but that is not so. Anyway, for any value of ##p## (whether correct or not) you now ask: if you toss your biased coin 5 times, what is the probability of getting at least 4 heads? Getting a 'head' in a toss is like an individual having DBP > 90, so that is really the same question as the one you are asked. Surely you have seen similar coin-tossing probability problems in the past?
 

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