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