The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent more than 90 days overdue.The historical records of the company show that over the past 8 years 13 percent of the accounts are delinquent.For this quarter,the auditing staff randomly selected 250 customer accounts.What is the probability that no more than 40 accounts will be classified as delinquent?
A)0.4207
B)0.9207
C)0.0793
D)0.4015
E)0.9015

Question

Quizplus · Accepted Answer

The problem can be approached using the binomial distribution formula or normal approximation to the binomial distribution. Given the large sample size (n = 250) and a known probability of success (p = 0.13), the normal approximation is suitable. The mean (μ) of the distribution is np = 250 * 0.13 = 32.5, and the standard deviation (σ) is sqrt(np(1-p)) = sqrt(250*0.13*0.87) ≈ 5.1. To find the probability of no more than 40 accounts being delinquent, we calculate the Z-score for 40.5 (for continuity correction) and use the standard normal distribution. The calculation yields a probability close to 0.9207, indicating that there's about a 92.07% chance no more than 40 accounts will be delinquent.

The Internal Auditing Staff of a Local Manufacturing Company Performs