1. Show that the function above satisfies the probabilities of a joint probability mass function.
2. Find E(X), E(Y), V(X), V(Y)
3. Find marginal probability distribution of X
4. Find the covariance and correlation.

On a night drive, a driver’s response time to any animal crossing the highway is normally distributed with a mean of 0.4 seconds and a standard deviation of 0.05 seconds. draw standard normal graphs and shade the regions clearly.

1. What is the probability that a response takes more than 0.5 seconds?

2. What is the probability that a response takes between 0.4 and 0.5 seconds?

3. What response time is exceeded 90% of the time?

A random sample of workers have been surveyed and data collected on how long it takes them to travel to work. The data are in this file.

1. Compute a 99% confidence interval for the expected time taken to travel to work. Show all your working and state any assumptions you need to make in order for your confidence interval to be valid.
2. Interpret the confidence interval

Question 1

The binomial formula has two parts. The first part of the binomial formula calculates the number of combinations of X successes. The second part of the binomial formula calculates the probability associated with the combination of success and failures. If N=4 and X=2, and p = .5, what is that probability from the second part of the formula?

Group of answer choices

.0625

.5

.3750

.1563

Question 2

Normal Distribution Problem. The birth weight is of newborn babies is approximately normally distributed with a mean of 3.39 kg and a standard deviation of .55kg. Note: my probabilities are exact probabilities. Solutions using the standard normal table will be close.

Low birth rate babies are strongly associated with infant mortality and other complications. A baby that weighs less than 2.5kg is considered low birthweight. What is the probability of a baby less than 2.5kg?

Group of answer choices

1 - .8944

.0125

.4472

.0528

Question 3

Normal Distribution Problem. The birth weight is of newborn babies is approximately normally distributed with a mean of 3.39 kg and a standard deviation of .55kg. Note: my probabilities are exact probabilities. Solutions using the standard normal table will be close.

What is the probability of a baby born weighing less than 3kg?

Group of answer choices

.7609

.5217

.2391

.2609

Question 4

Normal Distribution Problem. Red Blood Cell Counts are expressed millions per cubic millimeter of whole blood. For healthy females, x has a approximately normal distribution with mu = 4.8 and sigma =.3. Note: my probabilities are exact probabilities. Solutions using the standard normal table will be close.

What is the red blood count at the 80^th percentile?

Group of answer choices

4.6427

5.1000

5.0525

.5244

Question 5

Normal Distribution Problem. The birth weight is of newborn babies is approximately normally distributed with a mean of 3.39 kg and a standard deviation of .55kg. Note: my probabilities are exact probabilities. Solutions using the standard normal table will be close.

What is the probability of a baby born weighing more than 5.00 kg?

Group of answer choices

.0017

1 - .4983

2.927

.4983

Two teaching methods and their effects on science test scores are being reviewed. A random sample of 16 16 students, taught in traditional lab sessions, had a mean test score of 80.8 80.8 with a standard deviation of 4.4 4.4 . A random sample of 14 14 students, taught using interactive simulation software, had a mean test score of 85.6 85.6 with a standard deviation of 5.1 5.1 . Do these results support the claim that the mean science test score is lower for students taught in traditional lab sessions than it is for students taught using interactive simulation software? Let μ1 μ 1 be the mean test score for the students taught in traditional lab sessions and μ2 μ 2 be the mean test score for students taught using interactive simulation software. Use a significance level of α=0.05 α = 0.05 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 2 of 4 : Compute the value of the t test statistic. Round your answer to three decimal places.

A 2009 Pew Research Survey of a nationally representative sample of 242 cell phone users, aged 16 to 17 years, found that 52% had talked on the phone while driving.

1. Find a 98% confidence interval for the percentage of all cell phone users aged 16 to 17 years old who have talked on the phone while driving.

2. Do these data provide evidence that a majority of 16 to 17 year old cell phone users talk on the phone while driving?

3. What sample size is needed for you to estimate the percentage of all cell phone users aged 16 to 17 years old who have talked on the phone while driving to within a margin of error of 0.5%?

1. A non-profit agency wants to estimate the proportion of adults who answer yes to the question, ‘Was there ever a time in your life when you did not have a place to live?' They want to estimate the percentage at the 95% level with a margin of error of nor more than 0.5%.

   1. Find a sample size that will guarantee their confidence interval will have a margin of error of less than 0.5%.

   2. Why is the sample size you found in part likely to be far bigger than necessary?

   3. To get a better estimate of the sample size they will need, they take a preliminary sample of 30 adults, of which 2 answered yes to the question. About what sample size should they realistically expect to need?

In Pennsylvania, 6,165,478 people voted in the election. Trump received 48.18% of the vote and Clinton recieved 47.46%. This doesn't add up to 100% because other candidates received votes. All together these other candidates received 100% - 48.18% - 47.46% = 4.36% of the vote.

Suppose we could select one person at random from the 6+ million voters in PA (note: PA is the common abbreviation for Pennsylvania). We are interested in the chance that we'd choose a Trump, Clinton, or Other voter.

Below is a probability table for the choice:

Suppose we take a simple random sample of ?=1500 n = 1500 voters from the 6+ million voters in PA. What is the expected number of Trump voters? What is the expected number of Clinton voters? To answer these questions, let ?1 T 1 be 1 if the first voter chosen for the sample voted for Trump and 0 if they voted for Clinton or another candidate. Let ?2 T 2 be 1 if the second voter chosen for the sample voted for Trump and 0 if they voted for Clinton or another candidate, and so on. Let's start with some very basic questions. Find: ?(?1000=1) P ( T 1000 = 1 ) ?(?1000=0) P ( T 1000 = 0 ) ?(?17) E ( T 17 )

The 2010 U.S. Census found the chance of a household being a certain size, the data is in the table. Size of household 1 Person (26.7%),2 Person (33.6%), 3 Person (15.8%), 4 Person (13.7%), 5 Person (6.3%), 6 Person (2.4%), 7 or more (1.5%) a) Is it unusual for a household to have six people in the family? b) If you did come upon many families that had six people in the family, what would you think?

In an effort to promote a new product, a marketing firm asks participants to rate the effectiveness of ads that varied by length (short, long) and by type of technology (static, dynamic, interactive). Higher ratings indicated greater effectiveness.

(a) Complete the F-table and make a decision to retain or reject the null hypothesis for each hypothesis test. (Assume experimentwise alpha equal to 0.05.)

State the decision for the main effect of length.

Retain the null hypothesis.

Reject the null hypothesis.    

State the decision for the main effect of technology.

Retain the null hypothesis.

Reject the null hypothesis.    

State the decision for the interaction effect.

Retain the null hypothesis.Reject the null hypothesis.    

(b) Based on the results you obtained, what is the next step?

Compute simple main effect tests for the significant interaction.

Compute pairwise comparisons for the technology factor.     

Compute pairwise comparisons for the length factor.

No further analysis is needed, because none of the effects are significant.

1. Below there is information for a 1 sample proportion hypothesis test:

H0:P= 0.75

HA:P> 0.75

α=0.01

1. 1. What is the smallest value of the test statistic that would provide enough evidence to reject the null hypothesis?
2. 2. If the sample size is 50, what is the value of σp?
3. 3. If the test statistic is 2 (sample size is still 50), what is the value of p?

A sample of n = 4 is selected from a population with µ = 50. After a treatment is administered to the individuals in the sample, the mean is found to be M = 55 and the variance is s2= 64.

a. For a two-tailed test, what is the null hypothesis using statistical notation?

b. For a two-tailed test, what is the alternative hypothesis using statistical notation?

c. What is the estimated standard error?

d. What is the value/s for tcrit for a two-tailed test with α = .05

e. What is the value/s for tobt?

f. Is the data statistically significant for a two-tailed test with α = .05? NOTE: simply writing that the effect is significant (e.g., only writing “reject the null”) without showing any work/calculations in parts a-e will result in point zero points.

g. What is estimated Cohen’s d?

h. if you increased the size of the sample, how will this affect the likelihood of rejecting the null hypothesis?

i. If the variance increases, how will this affect the likelihood of rejecting the null hypothesis?

1. Statistics at a certain college has been historically taught at two times: at 8 am and at 4 pm.

A random sample of 150 students that took the morning class results in a mean score of 81.2 points, with a standard deviation of 18 points, and a random sample of 150 students that took the afternoon class results in a mean score of 76.4 points and a standard deviation of 21 points. For this problem, assume that the sample sizes are large enough so that the sample standard deviations (S) are good approximations for the unknown population standard deviations (σ).

a. Compute a 95% confidence interval for the mean score for all students taking statistics at 8 am.

b. Compute a 95% confidence interval for the mean score for all students taking statistics at 4 pm.

c. Based on the confidence intervals, is there strong evidence to support the claim that the morning classes do better in statistics? Explain.

2. A poorly written research paper states a confidence interval for the mean reaction time to an experiment as 83.6 ± 11.515 seconds, but forgot to mention what the confidence level was. However, the paper did say that the population standard deviation is σ = 35, and the sample size was n = 25.

a. What was the confidence level used for the confidence interval stated in the paper? b. Using the same sample results, how could you lower the margin of error to below 10 seconds?

Find the monthly payment needed to amortize principal and interest for each fixed-rate mortgage for a $220,000 at 4.5% interest for 30 years.

1b The Earth is structured in layers: crust, mantle, and core. A recent study was conducted to estimate the mean depth of the upper mantle in a specific farming region in California. Twenty-six, n = 26 sample sites were selected at random from a normally distributed population of approximately N = 1598 sites, and the depth of the upper mantle was measured using changes in seismic velocity and density. The sample mean was 127.5 km and the sample standard deviation was 21.3 km. Suppose the depth of the upper mantle is normally distributed. Find a 90% confidence interval for the true mean depth of the upper mantle in this farming region.

1. Determine the prerequisites have been met (3 pts):

• Random sample:
• n ≤ 0.05 of N:
• the population is normally distributed or n ≤ 30:

2. Construct the confidence interval (4 pts):

3. Interpret the confidence interval, in a statement (1 pt):

1c According to the U.S. Fire Administration, approximately N = 25,000 fires are caused by fire-works each year in the United States. Despite numerous public warnings against the use of fireworks, the home property damage due to these fires is enormous. In a random sample of n = 25 fires due to fireworks, the resulting mean property damage (in dollars) was 860.75 with a standard deviation of 350.50. Assume the underlying distribution of property damage due to these fires is normal. Find a 99% confidence interval for the true mean property damage due to a fire caused by fireworks.

1. Determine the prerequisites have been met (3 pts):

• Random sample:
• n ≤ 0.05 of N:
• the population is normally distributed or n ≤ 30:

2. Construct the confidence interval (4 pts):

3. Interpret the confidence interval, in a statement (1 pt):