Question

Dottie's Tax Service specialises in federal tax returns for professional clients such as, Physicians, Dentists, Accountants and Lawyers. A recent audit by the IRS of the returns she prepared indicated that an error was made on 7 percent of the returns she prepared last year. Assuming this rate continues into this year and she prepares 80 returns, what is the probability that she makes errors on:

(a) More than six returns?

(b) At least six returns?

(c) Exactly six returns?

Answer 1

Here, n = 80, p = 0.07, (1 - p) = 0.93 and x = 5
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

a)
P(X <= 6) = (80C0 * 0.07^0 * 0.93^80) + (80C1 * 0.07^1 * 0.93^79) + (80C2 * 0.07^2 * 0.93^78) + (80C3 * 0.07^3 * 0.93^77) + (80C4 * 0.07^4 * 0.93^76) + (80C5 * 0.07^5 * 0.93^75) + (80C6 * 0.07^6 * 0.93^74)
P(X <= 6) = 0.003 + 0.0181 + 0.0539 + 0.1055 + 0.1528 + 0.1748 + 0.1645
P(X <= 6) = 0.6726

P(X > 6) = 1 - 0.6726 = 0.3274

b)
P(X <= 5) = (80C0 * 0.07^0 * 0.93^80) + (80C1 * 0.07^1 * 0.93^79) + (80C2 * 0.07^2 * 0.93^78) + (80C3 * 0.07^3 * 0.93^77) + (80C4 * 0.07^4 * 0.93^76) + (80C5 * 0.07^5 * 0.93^75)
P(X <= 5) = 0.003 + 0.0181 + 0.0539 + 0.1055 + 0.1528 + 0.1748
P(X <= 5) = 0.5081

P(X >= 6) = 1 - 0.5081 = 0.4919

c)
P(X = 6) = 80C6 * 0.07^6 * 0.93^74
P(X = 6) = 0.1645