James placed a $25 bet on red and a $5 bet on the number 33 (which is black) on a standard 00 roulette wheel. If the ball lands in a red space, he wins $25 on his 'red' bet but loses $5 on his '33' bet- so he wins $20. If the ball lands on the number 33, he loses $25 on his 'red' bet but wins $175 on his '33' bet: He wins $150 If the ball lands on a space that isn't red and isn't 33 he loses both bets, so he loses $30. So for each spin; he either wins $150, wins $20, or loses $30. The probability that he wins $150 is 1/38 or .0263 The probability that he wins $20 is 18/38 or .4737 The probability that he loses $30 is 19/38 or .5000 Let X = the profit that James makes on the next spin. P(X =x) x P(X x) 2. P(X = x) MxP(X x) = -$1.581 x a2 Ex2 P(X x) - 42 1231.23 1.5812 150 .0263 591.75 3.945 1 20 .4737 189.48 9.474 1 1228.73044 - 30 .5000 450.00 -15.000 Σ 1.0000 1231.23 -1.581 V1228.73044 35.053 If you play 2500 times, and Let, the mean winnings (or losses)per game. The expected value of , called uz is The standard deviation of 3, called oz is This means that for 2500 games, you expect your winnings (or loss) to be 2500 times your average loss per game, which is Of course, you won't always win (or lose) this amount. If many people played independently of each other, each using this same betting strategy, we'd see different profits. theorem tells me that these winnings/losses will be Normally distributed. The Multiplying the value found for og by 2500 (the number of games) we find that standard deviation in profit for 2500 games is Using the information from MAC25, we can calculate the probability of making money (winnings > 0) when playing SHOW YOUR WORK OR CALCULATOR INPUT 2500 to be ... The probability of having a profit over $1000 (win an average of $0.40 per game for 2500 games) is SHOW YOUR WORK OR CALCULATOR INPUT This means that we'd only expect one of every people repeating this bet 2500 times to profit, and we'd only expect one out of every to make over $1000. and the probability of losing over $6000 is The probability of losing over $3000 is SHOW YOUR WORK OR CALCULATOR INPUT There is a lesson to be learned with this problem (and the one on the next page). If you gamble a little, playing a game where the odds are against you, you may get lucky and have a profitable hour. You might even have a profitable day or weekend if you play for a few hours each day. If you do this repeatedly, however, the probability of having a profit over the course of many (in this example, 2500) games is very small.

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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USING TI-83 or TI-84 for the calculation Using both Page 1 and Page 2 to answer each questions

I NEED HELP on answering both sheets

 

James placed a $25 bet on red and a $5 bet on the number 33 (which is black) on a standard 00 roulette wheel.
If the ball lands in a red space, he wins $25 on his 'red' bet but loses $5 on his '33' bet- so he wins $20.
If the ball lands on the number 33, he loses $25 on his 'red' bet but wins $175 on his '33' bet: He wins $150
If the ball lands on a space that isn't red and isn't 33 he loses both bets, so he loses $30.
So for each spin; he either wins $150, wins $20, or loses $30.
The probability that he wins $150 is 1/38 or .0263
The probability that he wins $20 is 18/38 or .4737
The probability that he loses $30 is 19/38 or .5000
Let X = the profit that James makes on the next spin.
P(X =x)
x P(X x)
2. P(X = x)
MxP(X x) = -$1.581
x
a2 Ex2 P(X x) - 42
1231.23 1.5812
150
.0263
591.75
3.945
1
20
.4737
189.48
9.474
1
1228.73044
- 30
.5000
450.00
-15.000
Σ
1.0000
1231.23
-1.581
V1228.73044
35.053
If you play 2500 times, and
Let,
the mean winnings (or losses)per game. The expected value of , called uz is
The standard deviation of 3, called oz is
This means that for 2500 games, you expect your winnings (or loss) to be 2500 times your average loss per game,
which is
Of course, you won't always win (or lose) this amount. If many people played
independently of each other, each using this same betting strategy, we'd see different profits.
theorem tells me that these winnings/losses will be Normally distributed.
The
Multiplying the value found for og by 2500 (the number of games) we find that standard deviation in profit for
2500 games is
Transcribed Image Text:James placed a $25 bet on red and a $5 bet on the number 33 (which is black) on a standard 00 roulette wheel. If the ball lands in a red space, he wins $25 on his 'red' bet but loses $5 on his '33' bet- so he wins $20. If the ball lands on the number 33, he loses $25 on his 'red' bet but wins $175 on his '33' bet: He wins $150 If the ball lands on a space that isn't red and isn't 33 he loses both bets, so he loses $30. So for each spin; he either wins $150, wins $20, or loses $30. The probability that he wins $150 is 1/38 or .0263 The probability that he wins $20 is 18/38 or .4737 The probability that he loses $30 is 19/38 or .5000 Let X = the profit that James makes on the next spin. P(X =x) x P(X x) 2. P(X = x) MxP(X x) = -$1.581 x a2 Ex2 P(X x) - 42 1231.23 1.5812 150 .0263 591.75 3.945 1 20 .4737 189.48 9.474 1 1228.73044 - 30 .5000 450.00 -15.000 Σ 1.0000 1231.23 -1.581 V1228.73044 35.053 If you play 2500 times, and Let, the mean winnings (or losses)per game. The expected value of , called uz is The standard deviation of 3, called oz is This means that for 2500 games, you expect your winnings (or loss) to be 2500 times your average loss per game, which is Of course, you won't always win (or lose) this amount. If many people played independently of each other, each using this same betting strategy, we'd see different profits. theorem tells me that these winnings/losses will be Normally distributed. The Multiplying the value found for og by 2500 (the number of games) we find that standard deviation in profit for 2500 games is
Using the information from MAC25, we can calculate the probability of making money (winnings > 0) when playing
SHOW YOUR WORK OR CALCULATOR INPUT
2500 to be ...
The probability of having a profit over $1000 (win an average of $0.40 per game for 2500 games) is
SHOW YOUR WORK OR CALCULATOR INPUT
This means that we'd only expect one of every
people repeating this bet 2500 times to profit,
and we'd only expect one out of every
to make over $1000.
and the probability of losing over $6000 is
The probability of losing over $3000 is
SHOW YOUR WORK OR CALCULATOR INPUT
There is a lesson to be learned with this problem (and the one on the next page). If you gamble a little, playing a
game where the odds are against you, you may get lucky and have a profitable hour. You might even have a
profitable day or weekend if you play for a few hours each day. If you do this repeatedly, however, the probability of
having a profit over the course of many (in this example, 2500) games is very small.
Transcribed Image Text:Using the information from MAC25, we can calculate the probability of making money (winnings > 0) when playing SHOW YOUR WORK OR CALCULATOR INPUT 2500 to be ... The probability of having a profit over $1000 (win an average of $0.40 per game for 2500 games) is SHOW YOUR WORK OR CALCULATOR INPUT This means that we'd only expect one of every people repeating this bet 2500 times to profit, and we'd only expect one out of every to make over $1000. and the probability of losing over $6000 is The probability of losing over $3000 is SHOW YOUR WORK OR CALCULATOR INPUT There is a lesson to be learned with this problem (and the one on the next page). If you gamble a little, playing a game where the odds are against you, you may get lucky and have a profitable hour. You might even have a profitable day or weekend if you play for a few hours each day. If you do this repeatedly, however, the probability of having a profit over the course of many (in this example, 2500) games is very small.
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