posted by nick on .
Q) 75% of the population in Guinea, Africa is infected with malaria. Suppose
a physician sees 6 patients in a given hour. Assuming these conditions are met, find the mean and standard deviation of the distribution.
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?
would i therefore need to be doing
sigma = square root of [np (1-p)]?
oh 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?
I 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)
np=mean (expected value) of number of infected patients out of 6.
√(npq)=s.d. of mean.