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### Several Probability and Statistics Techniques

```
Date: 04/13/98 at 01:59:01
From: Gareth Evans
Subject: Probability

1) An investment corporation is formed with enough capital to finance
9 independent ventures. If the chance of being successful on any
one venture is 25%, what are the corporation's chances of having:

a) exactly one successful venture?
b) at least one successful venture?
c) all nine ventures unsuccessful?

I have no idea how to approach this question.

2) A safety supervisor at a factory believes the average number of
accidents per month to be 3.5. What is the probability of:

a) exactly 2 accidents next month?
b) three or more accidents?

I also have no idea with this one.
```

```
Date: 04/13/98 at 07:24:27
From: Doctor Anthony
Subject: Re: Probability

The first question is binomial probability with:

n = 9
p = probability of success = 1/4, and
q = probability of failure = 3/4

The formula for r successes in n trials is P(r) = C(n,r)p^r*q^(n - r),
where C(n,r) is the binomial coefficient:

n!
----------
r!(n - r)!

Thus, the probability of exactly one success is:

Prob(exactly 1 success) = C(9,1)(1/4)(3/4)^8 = 0.225254

The probability of at least one successful venture includes
1, 2, 3, ... through 9 successes, and is best calculated from:

1 - P(0) = 1 - (3/4)^9 = 0.924915

The probability that all nine ventures are unsuccessful is:

P(0) = (3/4)^9 = 0.075085

Question 2 is a Poisson probability with mean 3.5 for a period of a
month. So:

P(0) =     e^(-3.5)  =  0.0301974
P(1) = 3.5/1 * P(0)  =  0.1056908
P(2) = 3.5/2 * P(1)  =  0.1849590
P(3) = 3.5/3 * P(2)  =  0.2157855

So the probability of exactly 2 accidents is 0.18496.

To find probability of 3 or more we calculate:

1 - [P(0) + P(1) + P(2)] = 1 - 0.320847
= 0.679153

-Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```

```
Date: 05/06/98 at 07:25:33
From: Gareth Evans
Subject: Re: Probability

Dear Dr. Maths,

I am currently studying probability and am having a bit of trouble
with the following questions due on Friday. I was wondering if you
would be able to help me. If you can, thank you very much. The
questions are:

1.) An office manager wishes to estimate the mean time required to
handle customer complaints. A sample of 38 complaints shows a mean
time of 28.7 minutes with a standard deviation of 12 minutes.

a) Construct a 90% confidence interval for the true mean time
required to handle customer complaints.
b) How large a sample should be taken, if the manager wishes to
estimate the mean time to handle a complaint to within +/- one
minute? Assume that the confidence interval is to be 90%.

2.) A hospital claims that the average length of stay is 5 days. A
study of the length of stay for a random sample of 25 patients
found the mean stay to be 6.2 days with a standard deviation of
1.2 days. Do these data present sufficient evidence to support the
hospital's claim? Use a 0.05 significance level.

3.) A survey was conducted to determine the proportion of registered
nurses in a particular state that are actively employed. A
random sample of 400 nurses selected from the state registry
showed 274 actively employed. Find a 95% confidence interval
estimate for the true proportion of registered nurses actively
employed in the state.

George Evans
```

```
Date: 05/06/98 at 11:26:56
From: Doctor Anthony
Subject: Re: Probability

In part b of the first question, the 90% confidence limits are given
by:

28.7 - xm
-----------  = +/- 1.645
12/sqrt(38)

where xm = population mean. Thus:

xm = 28.7 +/- 12/sqrt(38)*1.645
= 28.7 +/- 3.20225

And so:

25.498 < xm < 31.902

For question (1a), you require:

12/sqrt(n)*1.645 = 1

sqrt(n) = 12*1.645  =  19.74

n = 389.66

n = 390

For the second question, find the z-statistic:

6.2 - 5       1.2 * 5
------------ = --------- = 5
1.2/sqrt(25)      1.2

This value must be checked against z = 1.645 for a one-tailed 5% test.
It is a VERY significant result, and we reject the null hypothesis
that the mean is 5 days.

For the third question, we have:

p = 274/400 = 0.685
q = 0.315
n = 400
pq/n = .0005394
sqrt(pq/n)  =  0.023226

Then the true proportion of nurses employed is:

.685 - pm
---------- = +/- 1.96
0.023226

pm = .685 +/- 0.023226*1.96

= .685 +/- 0.04552

So:

0.63948 < pm < 0.7305

This is the 95% confidence interval for the proportion of nurses
actively employed in the state.

-Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability
High School Statistics

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