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September 3, 2014

September 3, 2014

Posted by **nick** on Thursday, August 9, 2012 at 1:52am.

a physician sees 6 patients in a given hour. Assuming these conditions are met, find the mean and standard deviation of the distribution.

Help please!

- statistics -
**MathMate**, Thursday, August 9, 2012 at 7:37amConditions:

1. probability remains constant throughout (i.e. for every one of the six patients.

2. the outcome is of type Bernoulli, i.e. yes/no, 1/0, etc. In this case, infected or not.

3. The number of patients is constant at 6 per hour.

4. Sampling can be assumed to be random and independent.

These conditions qualify the distribution as binomial, with parameters n=6, p=0.75 (probability of infection), and q=1-p=0.25 (probability of non-infection.

Can you find the mean and standard deviation (or variance) in terms of n, p and q?

- statistics -
**nick**, Friday, August 10, 2012 at 12:06amwould i therefore need to be doing

sigma = square root of [np (1-p)]?

- statistics -
**nick**, Friday, August 10, 2012 at 12:17amoh nevermind, i get what you are saying :)

would you recommend that i do two different groups of mean and variance,

one group is for having malaria,

the other for not?

- statistics -
**MathMate**, Friday, August 10, 2012 at 10:29pmI do not recommend doing the same thing both ways because this could cause confusion and consequent errors.

I suggest you document clearly. For example,

p=0.75=proportion of population infected

n=sample size (6 patients per hour)

so

np=mean (expected value) of number of infected patients out of 6.

√(npq)=s.d. of mean.

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