When an automobile is stopped by a roving safety patrol, each tire is checked for tire wear, and each headlight is checked to see whether it is properly aimed. Let X denote the number of headlights that need adjustment, and let Y denote the number of defective tires.

(a) If X and Y are independent with p_X(0) = 0.5, p_X(1) = 0.3, p_X(2) = 0.2, and p_Y(0) = 0.1, p_Y(1) = 0.2, p_Y(2) = p_Y(3) = 0.05, p_Y(4) = 0.6, display the joint pmf of (X, Y) in a joint probability table.

(b) Compute P(X ≤ 1 and Y ≤ 1) from the joint probability table.
P(X ≤ 1 and Y ≤ 1) =  

Does P(X ≤ 1 and Y ≤ 1) equal the product P(X ≤ 1) · P(Y ≤ 1)?

YesNo    

(c) What is P(X + Y = 0) (the probability of no violations)?
P(X + Y = 0) =  

(d) Compute P(X + Y ≤ 1).
P(X + Y ≤ 1) =

Question 4. The probability that an individual will suffer from a bad reaction of a given serum is .003. Out of 10000 individuals, compute the following probabilities using both the Binomial and Poisson models.

Part 1. P (X = 0) where X is Binomial and then Poisson.

Part 2. P (X > 10) where X is Binomial and then Poisson.

In the Focus Problem at the beginning of this chapter, a study was described comparing the hatch ratios of wood duck nesting boxes. Group I nesting boxes were well separated from each other and well hidden by available brush. There were a total of 489 eggs in group I boxes, of which a field count showed about 266 hatched. Group II nesting boxes were placed in highly visible locations and grouped closely together. There were a total of 798 eggs in group II boxes, of which a field count showed about 262 hatched.

(a) Find a point estimate p̂1 for p1, the proportion of eggs that hatch in group I nest box placements. (Round your answer to three decimal places.) p̂1 = Find a 90% confidence interval for p1. (Round your answers to three decimal places.)

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(b) Find a point estimate p̂2 for p2, the proportion of eggs that hatch in group II nest box placements. (Round your answer to three decimal places.) p̂2 = Find a 90% confidence interval for p2. (Round your answers to three decimal places.)

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(c) Find a 90% confidence interval for p1 − p2. (Round your answers to three decimal places.)

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Test the claim that for the population of statistics final exams, the mean score is 73 using alternative hypothesis that the mean score is different from 73. Sample statistics include n=18, x¯¯¯=74, and s=16. Use a significance level of α=0.01. (Assume normally distributed population.)

The test statistic is

The positive critical value is

The negative critical value is

The conclusion is A. There is sufficient evidence to reject the claim that the mean score is equal to 73. B. There is not sufficient evidence to reject the claim that the mean score is equal to 73.

A car service charges customers a flat fee per ride (which is higher during rush hour traffic) plus charges for each minute and each mile. Suppose that, in a certain metropolotian area during rush hour, the flat fee is $3, the cost per minute is $0.20, and the cost per mile is $1.20. Let x be the number of minutes and y the number of miles. At the end of a ride, the driver said that the passenger owed $20.60 and remarked that the number of minutes was five times the number of miles. Find the number of minutes and the number of miles for this trip.

A. Complete the equation that represents the total cost of the ride. ___=20.60

B. Complete the equation that represents the relationship between the number of minutes and number of miles. __=0

C. Find the number of minutes and number of miles for this trip.

A survey was taken involving the Spring 2011 Math 1530 students at ETSU. One of the questions asked was "When a person has a disease that cannot be cured, do you think doctors should be allowed by law to end the patient’s life by some painless means if the patient and his/her family request it?”  

Here is a table of the responses

Read the questions carefully. Word order makes a difference.

Question 9 (1 point)

If we pick one Spring 2011 Math 1530 student at random, what is the probability that we get a male student?

PERCENT ANSWER WITH TWO PLACES AFTER THE DECIMAL.

(1 point) A random sample of 100100 observations from a population with standard deviation 13.8313.83 yielded a sample mean of 92.392.3.

1. Given that the null hypothesis is μ=90μ=90 and the alternative hypothesis is μ>90μ>90 using α=.05α=.05, find the following:
(a) Test statistic ==  
(b) P - value:  
(c) The conclusion for this test is:

A. Reject the null hypothesis
B. There is insufficient evidence to reject the null hypothesis
C. None of the above

2. Given that the null hypothesis is μ=90μ=90 and the alternative hypothesis is μ≠90μ≠90 using α=.05α=.05, find the following:
(a) Test statistic ==  
(b) P - value:  
(c) The conclusion for this test is:

A. Reject the null hypothesis
B. There is insufficient evidence to reject the null hypothesis
C. None of the above

A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let X denote the number of hoses being used on the self-service island at a particular time, and let Y denote the number of hoses on the full-service island in use at that time. The joint pmf of X and Y appears in the accompanying tabulation.

(a) What is P(X = 1 and Y = 1)?
P(X = 1 and Y = 1) = _____

(b) Compute P(X ≤ 1 and Y ≤ 1).
P(X ≤ 1 and Y ≤ 1) = ______

Compute the probability of this event.
P(X ≠ 0 and Y ≠ 0) =  

(d) Compute the marginal pmf of X.

Compute the marginal pmf of Y.

Using p_X(x), what is P(X ≤ 1)?
P(X ≤ 1) = ________

Suppose two people (let’s call them Julio and Karina) agree to meet for lunch at a certain restaurant, each person’s arrival time, in minutes after noon, follows a normal distribution with mean 30 and standard deviation 10. Assume that they arrive independently of each other and that they agree to wait for 15 minutes. If each person agrees to wait exactly fifteen minutes for the other before giving up and leaving.

h) Report the probability distribution of the difference (not absolute difference) in the arrival times of Julio and Karina. [Hint: You might let Tj represent Julio’s arrival time and Tk represent Karina’s arrival time, both in minutes after noon. Use what you know about normal distributions to specify the probability distribution of the difference D = Tj – Tk.]

i) Use appropriate normal probability calculations to determine the probability that the two people successfully meet. Also report the values of the appropriate z-scores. [Hint: First express the probability that they successfully meet in terms of the random variable D.]

j) Now let m represent the number of minutes that both people agree to wait, where m can be any real number. Determine the value of m so the probability of meeting is .9

k) Now suppose that Julio and Karina can only afford to wait for 15 minutes, but they want to have at least a 90% chance of successfully meeting. Continue to assume that their arrival times follow independent normal distributions with mean 30 and the same SD as each other. Determine how small that SD needs to be in order to meet their criteria. (As always, show your work.)

1 point) It is necessary for an automobile producer to estimate the number of miles per gallon achieved by its cars. Suppose that the sample mean for a random sample of 100100 cars is 28.228.2 miles and assume the standard deviation is 3.23.2 miles. Now suppose the car producer wants to test the hypothesis that μμ, the mean number of miles per gallon, is 29.829.8 against the alternative hypothesis that it is not 29.829.8. Conduct a test using α=.05α=.05 by giving the following:

(a)    positive critical zz score    

(b)    negative critical zz score    

(c)    test statistic    

The final conclustion is

A. We can reject the null hypothesis that μ=29.8μ=29.8 and accept that μ≠29.8μ≠29.8.
B. There is not sufficient evidence to reject the null hypothesis that μ=29.8μ=29.8.

The number of eggs that a female house fly lays during her lifetime is normally distributed with mean 840 and standard deviation 116. Random samples of size 82 are drawn from this population, and the mean of each sample is determined. What is the probability that the mean number of eggs laid would differ from 840 by less than 30? Round your answer to four decimal places.

If samples of size 39 are taken from a bimodal population, the distribution of sample means will be approximately normal. How can I be so sure of this?

A. The Law of Large Numbers says so
B. The Central Limit Theorem says so
C. The data is normal because the problem says so
D. It is a basic property of probability

You may need to use the appropriate appendix table to answer this question.

Alexa is the popular virtual assistant developed by Amazon. Alexa interacts with users using artificial intelligence and voice recognition. It can be used to perform daily tasks such as making to-do lists, reporting the news and weather, and interacting with other smart devices in the home. In 2018, the Amazon Alexa app was downloaded some 2,800 times per day from the Google Play store.† Assume that the number of downloads per day of the Amazon Alexa app is normally distributed with a mean of 2,800 and standard deviation of 860.

(a)

What is the probability there are 2,100 or fewer downloads of Amazon Alexa in a day? (Round your answer to four decimal places.)

(b)

What is the probability there are between 1,400 and 2,600 downloads of Amazon Alexa in a day? (Round your answer to four decimal places.)

(c)

What is the probability there are more than 3,100 downloads of Amazon Alexa in a day? (Round your answer to four decimal places.)

(d)

Suppose that Google has designed its servers so there is probability 0.02 that the number of Amazon Alexa app downloads in a day exceeds the servers' capacity and more servers have to be brought online. How many Amazon Alexa app downloads per day are Google's servers designed to handle? (Round your answer to the nearest integer.)

downloads per day

A survey found that​ women's heights are normally distributed with mean 62.6 in. and standard deviation 2.2 in. The survey also found that​ men's heights are normally distributed with a mean 67.3 in. and standard deviation 2.8. Complete parts a through c below. a. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 8 in. and a maximum of 6 ft 2 in. Find the percentage of women meeting the height requirement. The percentage of women who meet the height requirement is ​%. ​(Round to two decimal places as​ needed.) b. Find the percentage of men meeting the height requirement. The percentage of men who meet the height requirement is ​%. ​(Round to two decimal places as​ needed.) c. If the height requirements are changed to exclude only the tallest​ 5% of men and the shortest​ 5% of​ women, what are the new height​ requirements? The new height requirements are at least in. and at most in. ​(Round to one decimal place as​ needed.)

Determine the margin of error for a 99​% confidence interval to estimate the population proportion with a sample proportion equal to 0.80 for the following sample sizes.

a. n=100             b. n=200             c. n=260