Question

Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean μ and standard deviation σ. Also, use the range rule of thumb to find the minimum usual value μ−2σand the maximum usual value μ+2σ.

n=1475​,

p=3/5

Answer 1

Given that, n = 1475 and p = 3/5 = 0.6

Mean and standard deviation of the Binomial distribution are,

$\mu = np =1475*0.6 = 885$

$\sigma=\sqrt { np(1-p)}=\sqrt {1475*0.6*(1-0.6)}= 18.8149$

Therefore,

$\mu - 2\sigma= 885 - (2*18.8149) = 885 - 37.6298 = 847.3702 \approx 847.37$

$\mu + 2\sigma= 885 + (2*18.8149) = 885 + 37.6298 = 922.6298 \approx 922.63$

The minimum usual value = 847.37

The maximum uaual value = 922.63