Answer:
(a) average calls = 5
(b) probability that there is exactly one call in 6 consecutive monts = 0.038
Step-by-step explanation:
Let event of a customer requiring help in a particular month = H
and event of a customer not requiring help in a particular month = ~H
Given
p= 0.01, therefore
Number of households, n = 500.
Binomial distribution:
x = number of households requiring help in a particular month
P(x,n,p) = C(x,n)*p^x*(1-p)^(n-x)
where
C(x,n) = n!/(x!(n-x)!) is the the number of combinations of x objects out of n
(a) Average number of households requiring help = np = 500*0.01 = 5
(b)
Probability that there are no calls requiring help in a particular month
P(0), q= C(0,n)*p^0(1-p)^(n-0)
= 1*1*0.99^500
= 0.006570483
Applying binomial probability over six months,
q = 0.006570483
n = 6
x = 1
P(x,n,q)
= C(x,n)*q^x*(1-q)^(n-x)
= 6!/(1!*5!) * 0.006570483^1 * (1-0.006570483)^5
= 0.038145
Therefore the probability that in 6 consecutive months there is exactly one month that no customer has called for help = 0.038
Rounding the result to the nearest tenth:
Answer: