The probability of a rare disease striking a described population is 0.003. A sample of 10000 was examined. Determine the expected no. suffering from the disease and thus find out the variance and the standard deviation for the above problem

Solution

Sample size n = 10000

P(a person suffering from the disease) = 0.003 = p

∴ expected number of people suffering from the disease

Mean = λ = 10000 × 0.003

            = 30

            = np = ?

Variance = np = 30

Standard deviation = √(np) = √?

                        =  √(30)

                        = 5.477