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.
Help please!
Conditions:
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)
so
np=mean (expected value) of number of infected patients out of 6.
√(npq)=s.d. of mean.