# please i need help in solving these questions

#### help101

##### New Member
hi all, please can anyone help me

1. The internal auditing staff of an 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 % of accounts are delinquent. For this quarter, the auditing staff randomly selected 25 customer accounts. What is the probability the no more than 40 accounts will be classified as delinquent?

2. A PH.D graduate has applied for a job with two universities, A and B. the graduate feels that she has a 60% chance of receiving an offer from university A, and a 50% chance of receiving an offer from university B. if she receives an offer from university B, she believes that she has an 80% chance of receiving an offer from university A:
- What the probability that both universities will make her an offer?
- What the probability that at least one university will make her an offer?

#### help101

##### New Member
question no.1 i didn't know to start
for the second question i solve it like this

p(A and B) = p(A) * p(A/B)= .6 * .8 = .48
p(A or B) = p(A) + p(B) - p(A and B) = .6 + .5 - .48 = .62 .... but i got zero

#### noetsi

##### Fortran must die
For this quarter, the auditing staff randomly selected 25 customer accounts. What is the probability the no more than 40 accounts will be classified as delinquent?
I would say the probability was zero because if you only surveyed 25 accounts you obviously can't find 40 accounts delinquent since 40 > 25

Are you sure you provided all the information from the question?

#### Dason

And for question 1 you might want to use the binomial distribution.

#### noetsi

##### Fortran must die
Well at least we found a case where p = o (in the original example)

#### help101

##### New Member
for 2 b
p(A or B) = p(A) + p(B) - p(A and B) = .6 + .5 - .40 = .7 right :\$

#### help101

##### New Member
no i know the binomial distribution.. and i tried to solve it and got this answer 90.15% but its wrong !!!

#### BGM

##### TS Contributor
The number looks close.

BTW why do you know it is wrong? You have the suggested solution?

#### BGM

##### TS Contributor
Yes I have verified that answer 92.07% in R once by normal approximation. Do you know how to do?