Question

Dottie's Tax Service specializes 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 13% of the returns she prepared last year. Assuming this rate continues into this year and she prepares 52 returns, what is the probability that she makes errors on:

1. More than 7 returns? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.)

2. At least 7 returns? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.)

3. Exactly 7 returns? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.)

Answer 1

We have here given , n = 52 , p = 0.13

np = 52 * 0.13 = 7.68 > 5

n(1-p) =52 * (1-0.13) = 56.32 > 5  

We can use here normal approximation to binomial

$\mu$=np=52*0.13=6.76

o = Vnp(1 - p) = 52 * 0.13* (1 – 0.13) = 2.43

a)  

the probability that she makes errors on: More than 7 returns

P[X>7]

=P[X>7.5]...................by using continuity correction

PGC-M, 7.5 – 6.76 2.43

P(Z>0.3051)

=1-P(Z<0.30561)

=1-0.6199.....................by using normal probability table.

=0.3801

b) the probability that she makes errors on At least 7returns

P(X >7)

=P[X>6.5]...........................using continuity correction.

P PC-H, 6.5 – 6.76 2 .43

P(Z>-0.1072)

=1-P(Z<-0.1072)

=1-0.4573.....................by using normal probability table.

=0.5427

c) P[X=7]

=

P(X >7) - P(X>7)

=0.5427-0.3801

=0.1626