5. Random pigeonholing

100 pigeons p1,…,p100 fly into 500 labelled holes h1,…,h500. Each pigeon picks a hole uniformly at random and independently from the choices of the other pigeons.

1. What is the probability that at least one hole contains at least 2 pigeons.
   Hint: The answer is approximately 0.9999758457295991
2. What is the probability that at least one hole contains at least 3 pigeons.
   Hint: The answer is approximately 0.4361298523736379.

The fill amount of bottles of a soft drink is normally distributed, with a mean of 2.0 liters and a standard deviation of 0.05 liter. If you select a random sample of 25 bottles, what is the probability that the sample mean will be:

A.) Between 1.99 and 2.0 liters

B.) Below 1.98 liters

C.) Greater than 2.01 liters

D.) The probability is 99% that the sample mean amount of soft drink will be at least how much?

E.) The probability is 99% that the sample mean amount of soft drink will be between which two amounts?

1. According to an educational report, the amount of time that students spend “off-task” (such as by checking their phones) during a one-hour lecture is approximately normally distributed with a mean of 3.2 minutes and a standard deviation of 2.7 minutes. An educator is interested in determining at the α = 0.05 level if the average amount of time his students spend on their phones while he is lecturing differs from the value given in the journal. During a particular class period in which he has 37 students, he noted that the average amount of time students spent on their phones was 4.2 minutes.

(a) State the null and alternative hypotheses for this test.

(b) Compute an appropriate test statistic.

(c) Determine the p-value for this test.

(d) State, in words, your conclusion.

2. Construct a 95% confidence interval for the mean amount of time students spend on their phones. Does this confidence interval support your conclusion from the hypothesis test in part (d)? Why or why not?

Consider a population having a standard deviation equal to 9.94. We wish to estimate the mean of this population.

(a) How large a random sample is needed to construct a 95% confidence interval for the mean of this population with a margin of error equal to 1? (Round your answer to the next whole number.)

The random sample is __________ units.

(b) Suppose that we now take a random sample of the size we have determined in part a. If we obtain a sample mean equal to 345, calculate the 95% confidence interval for the population mean. What is the interval’s margin of error? (Round your answers to the nearest whole number.)

The 95% confidence interval is            [ , ] .

Margin of error = ____________

Playing Roulette In the game of roulette, a wheel consists of 38 slots numbered 0, 00, 1, 2, . . . , 36. To play the game, a ball is spun around the wheel and is allowed to fall into on of the numbered slots. If the number of the slot the ball falls into matches the number you select, you win $35; otherwise you lose $1.

1. Construct a probability distribution for the random variable X, the winnings of each spin.

2. Determine the mean and standard deviation of the random variable X.

3. Suppose that you play the game 100 times, simulate at least 5 possible outcomes. Describe the pattern.

4. Imagine a person who plays the game 1000 times a day, for 365 days. Simulate such a scenario. What is the frequency distribution of that person’s winnings and losings?

5. Suppose that you play the game 1000 times. Describe the sampling distribution of the mean amount won per game.

6. What is the probability of being ahead after playing the game 100 times?

7. What is the probability of being ahead after playing the game 200 times?

8. What is the probability of being ahead after playing the game 1000 times?

9. Based on your investigation, what lesson does this teach you? Write an essay to inform your gambler friend.

Hint: To find the mean and standard deviation:

x <- c(35, -1)
p <- c(1/38, 37/38)

(mu.X <- sum(x*p))

## [1] -0.05263158

(sigma.X <- sqrt(sum((x - mu.X)^2*p)))

## [1] 5.762617

We simulate two possible outcomes if a person plays the game 100 times.

sample(x, 100, replace = TRUE, p)

##   [1] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
##  [24] -1 -1 -1 -1 35 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
##  [47] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
##  [70] -1 35 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 35 -1 -1
##  [93] -1 -1 -1 -1 -1 -1 -1 -1

sample(x, 100, replace = TRUE, p)

##   [1] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
##  [24] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 35 -1 -1 -1 -1 -1 -1 -1 -1
##  [47] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
##  [70] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
##  [93] -1 -1 -1 -1 -1 -1 -1 -1

Repeat this simulation many times and describe the pattern.

Imagine a person who plays the game n=100n=100 times a day, for 365 days. We examine the sum:

day <- 365
daily.sum <- replicate(day,
                       sum(sample(x, 100, replace = TRUE, p)))
(tb <- table(daily.sum))

## daily.sum
## -100  -64  -28    8   44   80  116  152  188 
##   27   63   78   84   62   33   14    2    2

plot(tb)

As the number of games nn increase, the sums follow a normal distribution N(nμ,nσ2)N(nμ,nσ2). With n=1000n=1000 and class width 36,

day <- 365
daily.sum <- replicate(day,
                       sum(sample(x, 1000, replace = TRUE, p)))
(tb <- table(daily.sum))

## daily.sum
## -568 -532 -460 -424 -388 -352 -316 -280 -244 -208 -172 -136 -100  -64  -28 
##    1    1    4    4    4   12   16   16   23   17   27   26   26   24   25 
##    8   44   80  116  152  188  224  260  296  332  368  440 
##   32   19   26   16   12   11    5    4    7    3    1    3

plot(tb/day/36)
curve(dnorm(x, 1000*mu.X, sqrt(1000)*sigma.X), add = TRUE)

The distribution of sample means has a standard deviation of σ/n−−√σ/n:

daily.mean <- daily.sum/1000
hist(daily.mean, probability = TRUE,
     xlim = c(-2, 2), ylim = c(0, 2.5))
curve(dnorm(x, mu.X, sigma.X/sqrt(1000)),
      xlim = c(-2, 2), add = TRUE)

Play 100 times. What is the probability of being ahead?

n1 <- 100
curve(dnorm(x, mu.X, sigma.X/sqrt(n1)), xlim = c(-2, 2),
      ylim = c(0, 2.5), main = "n=100", ylab = "pdf")
abline(v = mu.X)
coord.x <- c(0, seq(0, 2, 0.01), 2)
coord.y <- c(0, dnorm(seq(0, 2, 0.01), mu.X, sigma.X/sqrt(n1)), 0)
polygon(coord.x, coord.y, density = 10)

1 - pnorm(0, mu.X, sigma.X/sqrt(n1))

## [1] 0.4636141

Play 200 times:

n2 <- 200
curve(dnorm(x, mu.X, sigma.X/sqrt(n2)), xlim = c(-2, 2),
      ylim = c(0, 2.5), main = "n=200", ylab = "pdf")
abline(v = mu.X)
coord.x <- c(0, seq(0, 2, 0.01), 2)
coord.y <- c(0, dnorm(seq(0, 2, 0.01), mu.X, sigma.X/sqrt(n2)), 0)
polygon(coord.x, coord.y, density = 10)

1 - pnorm(0, mu.X, sigma.X/sqrt(n2))

## [1] 0.4486139

Play 1000 times:

n3 <- 1000
curve(dnorm(x, mu.X, sigma.X/sqrt(n3)), xlim = c(-2, 2),
      ylim = c(0, 2.5), main = "n=1000", ylab = "pdf")
abline(v = mu.X)
coord.x <- c(0, seq(0, 2, 0.01), 2)
coord.y <- c(0, dnorm(seq(0, 2, 0.01), mu.X, sigma.X/sqrt(n3)), 0)
polygon(coord.x, coord.y, density = 10)

1 - pnorm(0, mu.X, sigma.X/sqrt(n3))

## [1] 0.3863597

A sample of size 18 will be drawn from a population with mean 4 and standard deviation 3.

(a) Is it appropriate to use the normal distribution to find probabilities for x?

(b) If appropriate find the probability that x will be greater than 3.

(c) If appropriate find the 20th percentile of x.

of nine executives in a business firm, four are married, three have never married, and two are divorced. three of the executives are to be selected for promotion. Let Y1 donate the number of married executives and Y2 donate the number of never married executives among the three selected for promotion. Assuming that three are randomly selected from nine available find the joint probability function of Y1 and Y2.

An ecologist hypothesizes that birds with shorter wing spans use wider tree branches. The ecologist captured male birds, measured their wing span and other characteristics in millimeters, and then marked and released them. During the ensuing winter, the ecologist repeatedly observed the marked birds as they foraged for food on tree branches. He noted the branch diameter on each occasion, and calculated the average branch diameter for each bird in centimeters. The measurement data are below. What can the ecologist conclude with an α of 0.01?

a) What is the appropriate statistic?
---Select--- na Correlation Slope Chi-Square
Compute the statistic selected above:

b) Compute the appropriate test statistic(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  ; test statistic =
Decision:  ---Select--- Reject H0 Fail to reject H0

c) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
effect size =  ;   ---Select--- na trivial effect small effect medium effect large effect

d) Make an interpretation based on the results.

There was a significant positive relationship between the wing span of the birds and branch diameter.There was a significant negative relationship between the wing span of the birds and branch diameter.    There was no significant relationship between the wing span of the birds and branch diameter.

A passcode can have 4 or 5 digits. Digits can be repeated and leading 0s are allowed. So, 1234 would be a 4 digit code that is different from 01234, which is a 5 digit code. How many different passcodes are possible?

A population has a mean of 400 and a standard deviation of 50. Suppose a sample of size 125 is selected and x is used to estimate μ.

a. What is the probability that the sample mean will be within +/- 4 of the population mean (to 4 decimals)?

b. What is the probability that the sample mean will be within +/- 11 of the population mean (to 4 decimals)?

A random sample of 11 items is drawn from a population whose standard deviation is unknown. The sample mean is x¯ = 920 and the sample standard deviation is s = 25. Use Appendix D to find the values of Student’s t.

(a) Construct an interval estimate of μ with 95% confidence. (Round your answers to 3 decimal places.)
  
The 95% confidence interval is from to

(b) Construct an interval estimate of μ with 95% confidence, assuming that s = 50. (Round your answers to 3 decimal places.)
  
The 95% confidence interval is from to

(c) Construct an interval estimate of μ with 95% confidence, assuming that s = 100. (Round your answers to 3 decimal places.)
  
The 95% confidence interval is from to

Men’s heights are normally distributed with a mean of 69.5 inches and a standard deviation of 2.4 inches.

1. What percent of men are taller than 5 feet?

2. What percent of men can be a flight attendant if you must be between 5’2” and 6’1”?

3. Find the 45th Percentile of men’s heights.

Let X and Y have the joint PDF

f(x) = { 1/2 0 < x + y < 2, x > 0, y > 0 ;

{ 0 elsewhere

a) sketch the support of X and Y

b) Are X and Y independent? Explain.

c) Find P(x<1 and y<1.5)

Since an instant replay system for tennis was introduced a a major tournament, men challenged 1438 referee calls, with the result that 422 of the calls were overturned. Women challenged 740 referee calls, and 212 of the calls were overturned. Use a .05 significance level to test the claim that men and women have equal success in challenging calls. Complete parts (a) through (c) below. a. test the claim using a hypothesis test. Consider the first sample to be the sample of male tennis players who challenged referee calls and the second sample to be the sample of female tennis players who challenged referee calls. What are the null and alternative hypotheses for the hypothesis test? Identify the test statistic. Z= (round to 2 decimal places as needed) Identify the P-value. P= (round to 3 decimal places as needed) What is the conclusion based on the hypothesis test? The P-value is Less (than/greater) than the significance level of alpha= .05, so (fail to reject/reject) the null hypothesis. There (is sufficient/is not sufficient) evidence to warrant rejection of the claim that women and men have equal success in challenging calls. B. Test the claim by constructing an appropriate confidence interval. The 95% confidence interval is x<(p1-p2)