Suppose that, on average, 1 person in 1000 makes a numerical error in preparing his or her income tax return. If 10,000 returns are selected at random and examined, find the probability that 6, 7, or 8 of them contain an error.

0.2657

Step-by-step explanation:

Given that:

Probability of Error (p) = 1/ 1000 = 0.001

Number of trials (n) = 10000

μ = (n * p) = 10000 * 0.001 = 10

Using the binomial distribution :

P(x, μ) = e^-μ (μ^x) / x!

for x = 6

P(6, 10) = e^-10 (10^6) / 6!

P(6, 10) = 0.0000453999 (1388.8888) = 0.06305541263112

for x = 7

P(7, 10) = e^-10 (10^7) / 7!

P(7, 10) = 0.0000453999 (1984.1269) = 0.09007916284731

for x = 8

P(8, 10) = e^-10 (10^8) / 8!

P(8, 10) = 0.0000453999 (2480.1587) = 0.11259895696413

P(6) + P(7) + P(8)

0.06305541263112 + 0.09007916284731+ 0.11259895696413

= 0.26573353244256

= 0.2657