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**deepu4****Guest**

Hi

Please help me with these probability questions

1. A manufacturer of small electric motors wishes to give a guarantee on all those motors that fail prematurely, but does not want to replace too many motors. If the lifetimes fo the electric motors are normally didtributed with a mean of 8 years and a standard deviation of 2 years, what guarantee should be offered so that not more than 5% of the motors are replaced?????

2. A normal distribution has mean 20 and standard deviation=3. If X is the mean of a random sample of size 49 taken from this diatribution then determine the following

a) P(18<X<22)

b) P(X>19.5)

c) P(X<21)

d) The number 'k' such that P(20-k<X<20+k) = 0.90

3. The weights fo ball bearings are normally distributed with mean 0.614 and standard deviation of 0.003

Determine the percentage of ball bearings with weights:

a) between 0.610 and 0.618

b) greater than 0.616

c) less than 0.607

d) Determine the new mean with standard deviation fixed so that the answer to question (b) is 25%

4. A small hire has three cars which it hires out by the day. Records show that the mean demand over the day is 2.7 cars per day. If it is assumed that the demand for cars follows a poisson distribution determine

the percentage of days on which

a) no cars are hired

b) 2 or 3 cars are hired

c) Demand outstrips the supply

d) IF the firm buys another car for hire determine the percentage reduction in the answer for question (c)

I GOT ANSWER FOR QUESTION 4 AS

a) 6.7%

b)46.5%

c)28.6%

d)13.73%

PLEASE TELL ME WHETHER THE ANSWER I GOT FOR QUESTION 4 IS CORRECT OR NOT

PLEASE ALSO TELL ME HOW TO SOLVE THE FIRST THREE PROBLEMS

THANKS SO MUCH IN ADVANCE

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Statistics is unlike other math because you can't figure it out without referring to sources, graphs, tables. Statistics means follow the rules you've been taught without any freedom of thought.

*Last edited by John E. Franklin (2006-02-28 00:54:05)*

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Try this perhaps, it is a new forum since September I think.

http://www.talkstats.com

**igloo** **myrtilles** **fourmis**

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**deepu4****Guest**

hi

thank you for your reply . :)

But i am not able to do question 1????????

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Okay, you've convinced me to restudy a little statistics.

I never liked the way they squared the differences from the mean in the standard deviation. I personally would have just taken the absolute value. Why not take absolute value and put to the power of 1.95, you know, it's just all man-made.

But I'll look into this...

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Let's first do an example comparing JohnsDeviation to Standard Deviation.

Two data points are given, 10 and 20.

JohnDeviation would be 5 because ((15-10) + (20-15))/2 = 5

StandardDeviation would be sqrt( ((15-10)^2+(15-20)^2) / 2 ) ,

which is also 5.

Now let's do another example.

One data point, the number 50. The deviation is zero for both ways I think.

Now another example.

Three data points: 20, 24, 28.

JohnDeviation is (4 + 4 + 0)/3 = 8/3, or 2.66666666

StandardDeviation is sqrt( (4^2 + 4^2 + 0^2)/3 ), or 3.265986324

Now let's do another example:

Five data points: 9, 19,19,19,19

JohnsDeviation = ((17 - 9)+(19-17)*4)/5, or 3.2

StandardDeviation is sqrt( (4*((19-17)^2) + (9-17)^2 )/5 ), or 4.

Interesting, huh?

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Next we have to learn what normal distribution is. I've used the bell shaped curve many years ago, but I'll do some reading and get back...

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Okay, I've got part of it. 1.96 is x/stdDev because you look it up in a table under the area of .475 because .475 is 2.5% less than 50% and therefore on both sides of the normal curve, you get 95% acceptable, twice as much as 47.5%.

So 1.96 is what the table says is x/stdDev so that only 5% will fail.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
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The reason I didn't integrate the curve is because I read about trying to integrate a Gaussian curve, and it is really hard to do. There are various approximations.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Okay, let's backup a little. I made a tiny mistake on the lookup table.

The reason requires you to visualize the normal curve on a time line.

And the time line goes from zero to infinite years, NOT NEGATIVE

INFINITY TO POSITIVE INFINITY years. So this first proves that this problem

is not totally valid in the first place. You can tell your professor that a small

percentage of the motors will breakdown before they were actually built!!!!!!

This is because the normal curve is symmetrical in both directions from the

8th year. So whoever made up the problem was an idiot!

Or it is just an approximation, in which case I will proceed, since the small

amount of the curve that goes in negative time is small enough to ignore, and

we should calculate that too, so we can tell the teacher about it.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

He's not really an idiot, but he shouldn't have called it a normal curve if it wasn't one. He/she should have said it only approximates a normal curve.

Anyway, at 4 stdDEVs before eight years, the motor was made, at time = zero.

At this point, (1 - 0.9999366) / 2 of the motors are already broken.

So tell your teacher that 0.00317% of the motors broke prior to being built!!!

As for the actual question at hand. At 8 years, half of the motors are kaput (broke),

but don't forget that if the manufacturer replaces a motor once, he still has to

replace the motor AGAIN, if it breaksdown a second time!!!! So this problem is

again really more complicated than the teacher is probably concerned with.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
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So anyway, you have to lookup .45 for the area in the table to get x/stdDEV,

this is because you should only be guaranteeing for a few years on the left side

of the bell curve until an area of 5% is reached of kaputs (brokens).

So lookup 45%, which is the area from mean to one side of interest.

Then multiply that value by 2 years to get the time prior to year 8.

So 8 - (2*number looked up) should be the guarantee, but the company will

really replace more because the some replacements are recursive (they happen two or more times within the warranty)

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

1.65 is the number looked up for 45.05%

Use 45.05%, not 44.95% in the table because you want to be on the

safe side, less than 5% break.

So 8 - 2*1.65 = 4.7 years.

So the guarantee should go for 4.7 years for under 5% of customers having breakdowns. But really remember, some will be replaced twice or more, so

the company loses out unless they say you only get one replacement, which is

a really bad idea. So if this was a real company, you would have to lower the

number even more, and offer a three year warranty, or do more math to

figure it out somehow. Good Luck to you, deepu4.

**igloo** **myrtilles** **fourmis**

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**deepu4****Guest**

Thank you so much

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