Question

A big problem for taxpayers has been the recent increase in fraudulent tax returns being filed with the IRS in the name of taxpayers. IRS believes that 15 percent of all taxpayers will be affected in 2016. Assuming the proportion of taxpayers who get “hacked” is 0.15, what is the probability that more than 40 in a random sample of 200 will have had this experience?

Answer 1

p = 0.15, n = 200

Mean, µ = n*p = 200 * 0.15 = 30

Standard deviation, σ = √(n*p*(1-p)) = √(200 * 0.15 * 0.85) = 5.0498

P(X > 40)

= P((X - µ)/σ > (40 - 30)/5.0498)

= P(z > 1.9803)

= 1 - P(z < 1.9803)

Using excel function:

= 1 - NORM.S.DIST(1.9803, 1)

= 0.0238

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Or using continuity correction:

P(X > 40) =

Using continuity correction :

= P(X > 40+0.5)

= P((X - µ)/σ > (40.5 - 30)/5.0498)

= P(z > 2.0793)

= 1 - P(z < 2.0793)

Using excel function:

= 1 - NORM.S.DIST(2.0793, 1)

= 0.0188