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madreJ [45]
3 years ago
14

Your friend Alex is planning a fundraising game night to raise money for a local children's hospital. She invites a few of your

friends over to test out some of the games using fake Monopoly money.
For her first game, you will roll a six sided die and you will win $5 if you roll a six, $1 if you roll an odd number, and $0 if you roll a 2 or a 4, you will also pay $1.50 for every roll. Before you decide to play you do a few calculations. You have watched several people play this game and decided to collect some data.

(25 points) Assuming that it is a fair six-‐sided die, what are your expected winnings if you play Alex’s game? Who has the advantage in this game? Show your work and explain your result in a complete sentence.
(25 points) You’ve been watching a few people play this game and have observed that in the last five rolls none
f the players have won any money; that is, all rolls have been 2’s and 4’s. If Alex was using a fair die what is the probability of rolling only 2’s and 4’s five times in a row? Show your work and convey your results in a complete sentence.
You begin to suspect that Alex’s die is unfair, and decide to collect data regarding the results from several players. You would like to determine if the data you have observed is plausible assuming the die is fair.

c) (25 points) Of the 100 rolls you observed, a 6 is rolled only eight times. Based on this data, construct and interpret a 95% confidence interval for the proportion of rolls that are a 6 using Alex’s die. Do you believe that Alex’s die was a fair die? Justify your response with complete sentences, based on the confidence interval you constructed.

The following table displays the all the data from the 100 rolls you observed.

Outcome Frequency
1 10
2 29
3 11
4 30
5 12
6 8
You show Alex your findings, and she is embarrassed. She would like to develop a dice game that is fairer for the guests at her fundraiser.

d) (25 points) Using the empirical data above, give advice to Alex on how she should set up a game using her dice where she can still make money for her fundraiser but it might be more enticing for the guests to play. Give her an example of a game that she can use. Justify your recommendation with an expected value calculation and explain its significance to the situation in complete sentences.
Mathematics
1 answer:
katrin [286]3 years ago
8 0

Step-by-step explanation:

(a) You win $5 if you roll a six, $1 if you roll an odd number, and $0 if you roll a 2 or a 4, and you pay $1.50 for every roll.  The expected value is the sum of each outcome multiplied by its probability.

E = (5.00)(1/6) + (1.00)(3/6) + (0)(2/6) + (-1.50)(1)

E = -0.167

You are expected to lose on average about $0.17 per roll, which means Alex has the advantage.

(b) The probability of rolling a 2 or 4 on a fair die is 2/6 or 1/3.  The probability of this happening five times is:

P = (1/3)⁵

P = 1/243

P ≈ 0.41%

There is approximately a 0.4% probability that a fair die will roll a 2 or 4 five times.

(c) The confidence interval for a proportion is:

CI = p ± ME

ME = CV × SE

The margin of error is the critical value times the standard error.

The critical value for 95% confidence is z = 1.960.

The standard error for a proportion is:

SE = √(pq/n)

Given p = 1/6, q = 5/6, and n = 100:

SE = √((1/6) (5/6) / 100)

SE = 0.037

So the confidence interval is:

CI = 1/6 ± (1.960) (0.037)

CI = 0.167 ± 0.073

0.094 < p < 0.240

Since the observed proportion of 0.08 is outside of this interval, we can conclude with 95% confidence that the die is not fair.

(d) Under the current game rules and die probabilities, the expected value is:

E = (5.00)(0.08) + (1.00)(0.33) + (0)(0.59) + (-1.50)(1)

E = -0.77

To make the game fairer, but to still give Alex the advantage so she can make money for her fundraiser, we need to change the rules of the game so that the expected value is less negative.

One simple way to do this is to pay players $2.00 if they roll a 2.

Now the expected value is:

E = (5.00)(0.08) + (1.00)(0.33) + (2.00)(0.29) + (0)(0.30) + (-1.50)(1)

E = -0.19

Now instead of expecting to lose on average $0.77 per roll, players can expect to lose on average $0.19 per roll.  This means they have a better chance of winning money, but Alex still has the advantage.

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Suppose that the amount of time T a customer spends in a bank is exponentially distributed with an average of 10 minutes. What i
Bad White [126]

Answer:

The probability that a customer will spend more than 15 minutes total in the bank, given that the customer has already waited over 10 minutes  is 0.6065.

Step-by-step explanation:

The random variable <em>T</em> is defined as the amount of time a customer spends in a bank.

The random variable <em>T</em> is exponentially distributed.

The probability density function of a an exponential random variable is:

f(x)=\lambda e^{-\lambda x};\ x>0

The average time a customer spends in a bank is <em>β</em> = 10 minutes.

Then the parameter of the distribution is:

\lambda=\frac{1}{\beta}=\frac{1}{10}=0.10

An exponential distribution has a memory-less property, i.e the future probabilities are not affected by any past data.

That is, <em>P</em> (<em>X</em> > <em>s</em> + <em>x</em> | <em>X</em> ><em> s</em>) = <em>P</em> (<em>X</em> > <em>x</em>)

So the probability that a customer will spend more than 15 minutes total in the bank, given that the customer has already waited over 10 minutes  is:

P (X > 15 | X > 10) = P (X > 5)

\int\limits^{\infty}_{5} {f(x)} \, dx =\int\limits^{\infty}_{5}  {\lambda e^{-\lambda x}} \, dx\\=\int\limits^{\infty}_{5}  {0.10 e^{-0.10 x}} \, dx\\=0.10\int\limits^{\infty}_{5}  {e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10}|^{\infty}_{5}\\=[-e^{-0.10 \times \infty}+e^{-0.10 \times 5}]\\=0.6065

Thus, the probability that a customer will spend more than 15 minutes total in the bank, given that the customer has already waited over 10 minutes  is 0.6065.

8 0
3 years ago
How many bitstrings of length 10 contain three consecutive 0’s or 4 consecutive 1’s? How many bitstrings of length 10 contain tw
Yuki888 [10]

Answer:

147 bitstrings.

Step-by-step explanation:

To start with, we will compute the number of bit strings that has 3 consecutive 0's.

For each of the consecutive 0's, they can start at either 1st, 2nd, 3rd or 4th positions (because there are eight positions)

If we begin from the 1st position, there will be strings in the form of 000xxxxx

The other positions can be anything, count= 25 =32.

If we begin from the 2nd position, there will be strings in the form of 1000xxxx.

Note that the 1st position must contain 1, otherwise there will be more than one count strings.

The remaining 4 positions can be anything, count= 24 = 16.

If we begin from the 3rd position, there will be strings in the form of x1000xxx.

Note that the 2nd position must contain 1, or there will be more than one count strings as in the first scenario.

The remaining 4 positions can be anything, count= 24=16.

If we begin from the 4th, 5th, or 6th position, it would be the same analysis.

Therefore,

total count= 32 + 16 + 16 + 16 + 16 +16 = 112.

From the analysis done so far, we have double counted these 5 strings: 00010000, 00010001, 00001000, 00011000, 10001000

So, the actual strings that contain 3 consecutive 0's is 112-5 = 107.

If we also calculate the number of strings with 4 consecutive 1's, we will have: 16 + 8 + 8 + 8 + 8 = 48.

Therefore, there are 8 strings that we have double counted and they are: 11110000, 11110001, 11111000, 01111000, 00011110, 00011111, 00001111, 10001111.

So, for our final answer, the total number of bit strings of length 8 that contain either three consecutive 0's or four consecutive 1's is 107 + 48 - 8= 147.

7 0
3 years ago
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