MATH SOLVE

2 months ago

Q:
# The probability that a random smoker will develop a sever lung condition during his or her lifetime is 0.3. we will choose a random sample of 120 smokers and let x be the number of these smokers that will develop a severe lung condition.a.in our sample, find the mean and standard deviation of the number of smokers that will develop a sever lung condition. give your answers to two decimal plac

Accepted Solution

A:

The mean is 36 and the standard deviation is 5.02.

The mean is given by

ΞΌ = np = 120*0.3 = 36.

The standard deviation is given by

Ο =Β β(n*p*(1-p)) =Β β(120*0.3*0.7) =Β β25.2 = 5.02.

The mean is given by

ΞΌ = np = 120*0.3 = 36.

The standard deviation is given by

Ο =Β β(n*p*(1-p)) =Β β(120*0.3*0.7) =Β β25.2 = 5.02.