Post a New Question


posted by .

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!

  • statistics -

    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 -

    would i therefore need to be doing
    sigma = square root of [np (1-p)]?

  • statistics -

    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?

  • statistics -

    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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question