Question

It is found that 60% of American victims of healthcare fraud are senior citizens. Suppose that I have randomly sampled 100 victims, and I am looking at the count, x, of how many of those victims were senior citizens. Find the following:

a. the mean, variance and standard deviation of the distribution

b. the probability that at least 50 are senior citizens

c. the probability that 75 of them are senior citizens

d. the probability that less than 55 of them are senior citizens

Answer 1

Part a)

Mean = n * P = ( 100 * 0.6 ) = 60
Variance = n * P * Q = ( 100 * 0.6 * 0.4 ) = 24
Standard deviation = = 4.899

Part b)

P ( X >= 50 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 50 - 0.5 ) =P ( X > 49.5 )

P ( X > 49.5 ) = 1 - P ( X < 49.5 )
Standardizing the value

Z = ( 49.5 - 60 ) / 4.899
Z = -2.14

P ( Z > -2.14 )
P ( X > 49.5 ) = 1 - P ( Z < -2.14 )
P ( X > 49.5 ) = 1 - 0.0162
P ( X > 49.5 ) = 0.9838

Part c)

P ( X = 75 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 75 - 0.5 < X < 75 + 0.5 ) = P ( 74.5 < X < 75.5 )

P ( 74.5 < X < 75.5 )
Standardizing the value

Z = ( 74.5 - 60 ) / 4.899
Z = 2.96
Z = ( 75.5 - 60 ) / 4.899
Z = 3.16
P ( 2.96 < Z < 3.16 )
P ( 74.5 < X < 75.5 ) = P ( Z < 3.16 ) - P ( Z < 2.96 )
P ( 74.5 < X < 75.5 ) = 0.9992 - 0.9985
P ( 74.5 < X < 75.5 ) = 0.0008

Part d)

P ( X < 55 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 55 - 0.5 ) = P ( X < 54.5 )

P ( X < 54.5 )
Standardizing the value

Z = ( 54.5 - 60 ) / 4.899
Z = -1.12

P ( X < 54.5 ) = P ( Z < -1.12 )
P ( X < 54.5 ) = 0.1314