Hi wta2k:
First thing, let's look at what information we're given in the
question:
* average lifespan = mu = 63,000 miles
* standard deviation = sigma = 10,000 miles
* replacement cutoff level = L = 50,000 miles
* lifespans are normally distributed
a) For the current pump design, what percentage of the company’s pumps
will have to be replaced at no charge to the consumer?
To answer this, we need to find out how many standard deviations
(sigma) the replacement cutoff level (L) is away from the average
lifespan (mu). This value is called the "z-value"
z = (L - mu)/sigma
= (50,000 - 63,000)/10,000
= -1.3
So, now we need to find what probability a value of z=-1.3 corresponds
to on the normal distribution table. In the following online normal
table
http://davidmlane.com/hyperstat/z_table.html
enter -1.3 in the "Z" box and click "Compute Area". This shows us that
the area *below* z=-1.3 (i.e., less than 50,000 miles) equates to
.0969 or 9.69%.
Therefore, 9.69% of the pumps will have to be replaced at no charge to
the consumer.
b. What percentage of the company’s pumps will fail at exactly 50,000
miles?
The number of pumps that will fail at exactly 50,000 miles will be
infinitesimal - that is, very, very, very small. The probability that
a pump would stop right on an exact value is almost zero.
c. What percentage of the company’s pumps will fail between 40,000 and
55,000 miles?
In order to answer this question, we have to compute the z-value for
both L=40,000 and L=50,000.
For L=50,000, z=-1.3 (see part a.)
For L=40,000, z=(L-mu)/sigma
=(40,000-63,000)/10,000
=-2.3
We know the area below z=-1.3 equates to .0969. Using the online table
provided above, we can see that the area below z=-2.3 equates to
.0107.
To find the area between 40,000 and 50,000, we merely subtract the two
probabilities:
.0969 - .0107 = .0862 (or 8.62%)
Therefore, 8.62% of the company's pumps will fail between 40,000 and
50,000 miles.
d. For what number of miles does the probability become 80% that a
randomly selected pump will no longer be effective?
To answer this question, you need to use the normal table "in
reverse".
Go to the table and enter .80 in the "Area below Z" box. Click
"Compute Z". This shows you that z = .8415. What does this mean? This
means that 80% of the pumps will last the average value *plus* (since
z is positive) .8415 times the standard deviation. Or in symbolic
terms:
mu + z*sigma = 63,000 + .8415 * 10,000
= 71,415
Therefore, at 71,415 miles, the the probability becomes 80% that a
randomly selected pump will no longer be effective.
Some additional links that might be of help:
Standard Deviation:
http://www.robertniles.com/stats/stdev.shtml
Statistics at Square One
http://bmj.com/collections/statsbk/2.shtml
I hope this helps you out. Thanks for using Google Answers.
websearcher-ga
Search Strategy:
*personal experience*
"normal distribution" table
://www.google.com/search?hl=en&ie=ISO-8859-1&q=%22normal+distribution%22+table
normal statistics standard deviation
://www.google.com/search?hl=en&lr=&ie=ISO-8859-1&safe=off&q=normal+statistics+standard+deviation |
,
0 Comments
)