### Scribe

Andrew here, and today we had two classes.

In the morning Mr. K reminded those who haven't tagged three sites must do so and there will also be a test next week. There were two problems we had to do and it took the whole class time to do those questions, mainly the second one.

1. 1200 light bulbs were tested for the number of hours of life. The mean life was 640 hours with a standard deviation of 50 hours. Assume the life in hours of light bulbs is normally distributed.

a) What percent of light bulbs lasted between 600 and 700 hours?

Z = (600 - M)/S = -0.8

Z = (700 - M)/S = 1.2

ShadeNorm(-0.8 , 1.2) = 0.673075

b) How many light bulbs could be expected to last between 600 and 700 hours?

0.673075 * 1200 = 807.69

Now the second question (which uses binomial distribution) took the morning class and the second class to answer, PLUS we had to do another question (which uses binomial distribution and normal approximation).

2. A shipment of 200 tires is known to include 40 defective tires. 5 tires are selected at random and each tire is replaced before the next one is selected.

What is the probability of getting:

a) at most 2 defective tires

First find the probability of getting a defective tire.

40 / 200 = 0.20

Also, make a list in L1. Your list should look like this:

Go to the home screen and press 2nd function VARS to go to the distribution menu and select the binompdf ( .

binompdf(# of trials (n) , probability of success (p))

binompdf( 5 , 0.20) -store-> L2

L2 should look like this:

To find the answer to a) you must go to the home screen and press 2nd function STAT and then go to the MATH menu and choose sum ( . SUM will add all the probabilities from the lowest value to the highest.

sum( List , low value , high value )

sum(L2 , 1 (refer to the 1st image; you are actually choosing ZERO on the L1) , 3 (you are choosing TWO because at most you have 2 defective tires.)) This syntax should look like this:

sum( L2 , 1 , 3 ) = 0.94208

b) What is the probability of getting at least 1 defective tire?

sum( L2 , 1 , 1 ) = 0.32768

1 - Ans = 0.67232 , 0r , sum( L2 , 2 , 6)

The sum( L2 , 2 , 6 ) answer seems easier to explain, so the defective tire can either be the 1st, 2nd, 3rd, 4th, or 5th tire which is why the lowest value is 2 and the highest value is 6.

c) What is the probability of getting 2 or 3 defective tires?

sum( L2 , 3 , 4 ) = 0.256

And finally, on to question 3.

3. Customs officers estimate that 10% of vehicles crossing the Canada-US border carry undeclared goods. One day they randomly searched 350 vehicles. What is the probability that 40 or more vehicles carried undeclared goods?

n = 350

p = 0.10

q = 0.90

Now there are 2 methods to solving this problem. ONE being the binomial distribution method, the other being the normal approximation method. There are also 2 methods within the normal approximation method. You can either use Z-scores or RAW data.

Binomial Distribution: Uses n and p.

binompdf( 350 , 0.10 ) -store-> L1

sum( L1 , 41 , 351 ) = 0.2088192368

Normal Approximation: Uses n, p, and q.

Method 1: Z-scores

(Note: I'm still unsure of what to actually do since I sort of forgot how to do this)

M = n*p = 35

S = square root (n*p*q) = 5.61248608

Z = (40 - M) / S = 0.8908708064

Z = (350 - M) / S = 56.1248608

normalcdf( low value , high value )

normalcdf( low Z , high Z ) = 0.1864992046

Method 2: Raw Data

normalcdf( 40 , 350 , M , S) = 0.1864992046

(Note: for some reason using binomial distribution and normal approximation gives different answers but are also really close, and I have no idea why)

WELL, that's about all we did in both classes. Anyone who reads this should make COMMENTS in case I've made mistakes.

HOMEWORK: page 130, questions 1-8

NEXT SCRIBE: Muuxi

In the morning Mr. K reminded those who haven't tagged three sites must do so and there will also be a test next week. There were two problems we had to do and it took the whole class time to do those questions, mainly the second one.

1. 1200 light bulbs were tested for the number of hours of life. The mean life was 640 hours with a standard deviation of 50 hours. Assume the life in hours of light bulbs is normally distributed.

a) What percent of light bulbs lasted between 600 and 700 hours?

Z = (600 - M)/S = -0.8

Z = (700 - M)/S = 1.2

ShadeNorm(-0.8 , 1.2) = 0.673075

b) How many light bulbs could be expected to last between 600 and 700 hours?

0.673075 * 1200 = 807.69

Now the second question (which uses binomial distribution) took the morning class and the second class to answer, PLUS we had to do another question (which uses binomial distribution and normal approximation).

2. A shipment of 200 tires is known to include 40 defective tires. 5 tires are selected at random and each tire is replaced before the next one is selected.

What is the probability of getting:

a) at most 2 defective tires

First find the probability of getting a defective tire.

40 / 200 = 0.20

Also, make a list in L1. Your list should look like this:

Go to the home screen and press 2nd function VARS to go to the distribution menu and select the binompdf ( .

binompdf(# of trials (n) , probability of success (p))

binompdf( 5 , 0.20) -store-> L2

L2 should look like this:

To find the answer to a) you must go to the home screen and press 2nd function STAT and then go to the MATH menu and choose sum ( . SUM will add all the probabilities from the lowest value to the highest.

sum( List , low value , high value )

sum(L2 , 1 (refer to the 1st image; you are actually choosing ZERO on the L1) , 3 (you are choosing TWO because at most you have 2 defective tires.)) This syntax should look like this:

sum( L2 , 1 , 3 ) = 0.94208

b) What is the probability of getting at least 1 defective tire?

sum( L2 , 1 , 1 ) = 0.32768

1 - Ans = 0.67232 , 0r , sum( L2 , 2 , 6)

The sum( L2 , 2 , 6 ) answer seems easier to explain, so the defective tire can either be the 1st, 2nd, 3rd, 4th, or 5th tire which is why the lowest value is 2 and the highest value is 6.

c) What is the probability of getting 2 or 3 defective tires?

sum( L2 , 3 , 4 ) = 0.256

And finally, on to question 3.

3. Customs officers estimate that 10% of vehicles crossing the Canada-US border carry undeclared goods. One day they randomly searched 350 vehicles. What is the probability that 40 or more vehicles carried undeclared goods?

n = 350

p = 0.10

q = 0.90

Now there are 2 methods to solving this problem. ONE being the binomial distribution method, the other being the normal approximation method. There are also 2 methods within the normal approximation method. You can either use Z-scores or RAW data.

Binomial Distribution: Uses n and p.

binompdf( 350 , 0.10 ) -store-> L1

sum( L1 , 41 , 351 ) = 0.2088192368

Normal Approximation: Uses n, p, and q.

Method 1: Z-scores

(Note: I'm still unsure of what to actually do since I sort of forgot how to do this)

M = n*p = 35

S = square root (n*p*q) = 5.61248608

Z = (40 - M) / S = 0.8908708064

Z = (350 - M) / S = 56.1248608

normalcdf( low value , high value )

normalcdf( low Z , high Z ) = 0.1864992046

Method 2: Raw Data

normalcdf( 40 , 350 , M , S) = 0.1864992046

(Note: for some reason using binomial distribution and normal approximation gives different answers but are also really close, and I have no idea why)

WELL, that's about all we did in both classes. Anyone who reads this should make COMMENTS in case I've made mistakes.

HOMEWORK: page 130, questions 1-8

NEXT SCRIBE: Muuxi

Andrew, this is a fantastic summary of today's class.

This was probably one of the more difficult lectures in the entire course. You summarized it, simply, succintly, elegantly and clearly.

Well done!!

Posted by Mr. Kuropatwa | 4/03/2006 11:38 PM

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