Suppose the heights of 18-year-old men are approximately normally distributed, with mean 68 inches and standard deviation 2 inches.

(a) What is the probability that an 18-year-old man selected at random is between 67 and 69 inches tall? (Round your answer to four decimal places.)


(b) If a random sample of twenty-nine 18-year-old men is selected, what is the probability that the mean height x is between 67 and 69 inches? (Round your answer to four decimal places.)


(c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this?

The probability in part (b) is much higher because the mean is larger for the x distribution.The probability in part (b) is much higher because the standard deviation is larger for the x distribution.    The probability in part (b) is much lower because the standard deviation is smaller for the x distribution.The probability in part (b) is much higher because the mean is smaller for the x distribution.The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.

Using R calculate the following properties of the Data Set given below:

(a) The average (mean) value for each of the four features (b)

(b) the standard deviation for each of the features

(c) repeat steps (a) and (b) but separately for each type of flower (d)

(d) draw four box plots, one for each feature, such that each figure shows three boxes, one for each type of flower. Properly label your axes in all box plots.

Data Set

{

5.1,3.5,1.4,0.2,Iris-setosa
4.9,3.0,1.4,0.2,Iris-setosa
4.7,3.2,1.3,0.2,Iris-setosa
4.6,3.1,1.5,0.2,Iris-setosa
5.0,3.6,1.4,0.2,Iris-setosa
5.4,3.9,1.7,0.4,Iris-setosa
4.6,3.4,1.4,0.3,Iris-setosa
5.0,3.4,1.5,0.2,Iris-setosa
4.4,2.9,1.4,0.2,Iris-setosa
4.9,3.1,1.5,0.1,Iris-setosa
5.4,3.7,1.5,0.2,Iris-setosa
4.8,3.4,1.6,0.2,Iris-setosa
4.8,3.0,1.4,0.1,Iris-setosa
4.3,3.0,1.1,0.1,Iris-setosa
5.8,4.0,1.2,0.2,Iris-setosa
5.7,4.4,1.5,0.4,Iris-setosa
5.4,3.9,1.3,0.4,Iris-setosa
5.1,3.5,1.4,0.3,Iris-setosa
5.7,3.8,1.7,0.3,Iris-setosa
5.1,3.8,1.5,0.3,Iris-setosa
5.4,3.4,1.7,0.2,Iris-setosa
5.1,3.7,1.5,0.4,Iris-setosa
4.6,3.6,1.0,0.2,Iris-setosa
5.1,3.3,1.7,0.5,Iris-setosa
4.8,3.4,1.9,0.2,Iris-setosa
5.0,3.0,1.6,0.2,Iris-setosa
5.0,3.4,1.6,0.4,Iris-setosa
5.2,3.5,1.5,0.2,Iris-setosa
5.2,3.4,1.4,0.2,Iris-setosa
4.7,3.2,1.6,0.2,Iris-setosa
4.8,3.1,1.6,0.2,Iris-setosa
5.4,3.4,1.5,0.4,Iris-setosa
5.2,4.1,1.5,0.1,Iris-setosa
5.5,4.2,1.4,0.2,Iris-setosa
4.9,3.1,1.5,0.1,Iris-setosa
5.0,3.2,1.2,0.2,Iris-setosa
5.5,3.5,1.3,0.2,Iris-setosa
4.9,3.1,1.5,0.1,Iris-setosa
4.4,3.0,1.3,0.2,Iris-setosa
5.1,3.4,1.5,0.2,Iris-setosa
5.0,3.5,1.3,0.3,Iris-setosa
4.5,2.3,1.3,0.3,Iris-setosa
4.4,3.2,1.3,0.2,Iris-setosa
5.0,3.5,1.6,0.6,Iris-setosa
5.1,3.8,1.9,0.4,Iris-setosa
4.8,3.0,1.4,0.3,Iris-setosa
5.1,3.8,1.6,0.2,Iris-setosa
4.6,3.2,1.4,0.2,Iris-setosa
5.3,3.7,1.5,0.2,Iris-setosa
5.0,3.3,1.4,0.2,Iris-setosa
7.0,3.2,4.7,1.4,Iris-versicolor
6.4,3.2,4.5,1.5,Iris-versicolor
6.9,3.1,4.9,1.5,Iris-versicolor
5.5,2.3,4.0,1.3,Iris-versicolor
6.5,2.8,4.6,1.5,Iris-versicolor
5.7,2.8,4.5,1.3,Iris-versicolor
6.3,3.3,4.7,1.6,Iris-versicolor
4.9,2.4,3.3,1.0,Iris-versicolor
6.6,2.9,4.6,1.3,Iris-versicolor
5.2,2.7,3.9,1.4,Iris-versicolor
5.0,2.0,3.5,1.0,Iris-versicolor
5.9,3.0,4.2,1.5,Iris-versicolor
6.0,2.2,4.0,1.0,Iris-versicolor
6.1,2.9,4.7,1.4,Iris-versicolor
5.6,2.9,3.6,1.3,Iris-versicolor
6.7,3.1,4.4,1.4,Iris-versicolor
5.6,3.0,4.5,1.5,Iris-versicolor
5.8,2.7,4.1,1.0,Iris-versicolor
6.2,2.2,4.5,1.5,Iris-versicolor
5.6,2.5,3.9,1.1,Iris-versicolor
5.9,3.2,4.8,1.8,Iris-versicolor
6.1,2.8,4.0,1.3,Iris-versicolor
6.3,2.5,4.9,1.5,Iris-versicolor
6.1,2.8,4.7,1.2,Iris-versicolor
6.4,2.9,4.3,1.3,Iris-versicolor
6.6,3.0,4.4,1.4,Iris-versicolor
6.8,2.8,4.8,1.4,Iris-versicolor
6.7,3.0,5.0,1.7,Iris-versicolor
6.0,2.9,4.5,1.5,Iris-versicolor
5.7,2.6,3.5,1.0,Iris-versicolor
5.5,2.4,3.8,1.1,Iris-versicolor
5.5,2.4,3.7,1.0,Iris-versicolor
5.8,2.7,3.9,1.2,Iris-versicolor
6.0,2.7,5.1,1.6,Iris-versicolor
5.4,3.0,4.5,1.5,Iris-versicolor
6.0,3.4,4.5,1.6,Iris-versicolor
6.7,3.1,4.7,1.5,Iris-versicolor
6.3,2.3,4.4,1.3,Iris-versicolor
5.6,3.0,4.1,1.3,Iris-versicolor
5.5,2.5,4.0,1.3,Iris-versicolor
5.5,2.6,4.4,1.2,Iris-versicolor
6.1,3.0,4.6,1.4,Iris-versicolor
5.8,2.6,4.0,1.2,Iris-versicolor
5.0,2.3,3.3,1.0,Iris-versicolor
5.6,2.7,4.2,1.3,Iris-versicolor
5.7,3.0,4.2,1.2,Iris-versicolor
5.7,2.9,4.2,1.3,Iris-versicolor
6.2,2.9,4.3,1.3,Iris-versicolor
5.1,2.5,3.0,1.1,Iris-versicolor
5.7,2.8,4.1,1.3,Iris-versicolor
6.3,3.3,6.0,2.5,Iris-virginica
5.8,2.7,5.1,1.9,Iris-virginica
7.1,3.0,5.9,2.1,Iris-virginica
6.3,2.9,5.6,1.8,Iris-virginica
6.5,3.0,5.8,2.2,Iris-virginica
7.6,3.0,6.6,2.1,Iris-virginica
4.9,2.5,4.5,1.7,Iris-virginica
7.3,2.9,6.3,1.8,Iris-virginica
6.7,2.5,5.8,1.8,Iris-virginica
7.2,3.6,6.1,2.5,Iris-virginica
6.5,3.2,5.1,2.0,Iris-virginica
6.4,2.7,5.3,1.9,Iris-virginica
6.8,3.0,5.5,2.1,Iris-virginica
5.7,2.5,5.0,2.0,Iris-virginica
5.8,2.8,5.1,2.4,Iris-virginica
6.4,3.2,5.3,2.3,Iris-virginica
6.5,3.0,5.5,1.8,Iris-virginica
7.7,3.8,6.7,2.2,Iris-virginica
7.7,2.6,6.9,2.3,Iris-virginica
6.0,2.2,5.0,1.5,Iris-virginica
6.9,3.2,5.7,2.3,Iris-virginica
5.6,2.8,4.9,2.0,Iris-virginica
7.7,2.8,6.7,2.0,Iris-virginica
6.3,2.7,4.9,1.8,Iris-virginica
6.7,3.3,5.7,2.1,Iris-virginica
7.2,3.2,6.0,1.8,Iris-virginica
6.2,2.8,4.8,1.8,Iris-virginica
6.1,3.0,4.9,1.8,Iris-virginica
6.4,2.8,5.6,2.1,Iris-virginica
7.2,3.0,5.8,1.6,Iris-virginica
7.4,2.8,6.1,1.9,Iris-virginica
7.9,3.8,6.4,2.0,Iris-virginica
6.4,2.8,5.6,2.2,Iris-virginica
6.3,2.8,5.1,1.5,Iris-virginica
6.1,2.6,5.6,1.4,Iris-virginica
7.7,3.0,6.1,2.3,Iris-virginica
6.3,3.4,5.6,2.4,Iris-virginica
6.4,3.1,5.5,1.8,Iris-virginica
6.0,3.0,4.8,1.8,Iris-virginica
6.9,3.1,5.4,2.1,Iris-virginica
6.7,3.1,5.6,2.4,Iris-virginica
6.9,3.1,5.1,2.3,Iris-virginica
5.8,2.7,5.1,1.9,Iris-virginica
6.8,3.2,5.9,2.3,Iris-virginica
6.7,3.3,5.7,2.5,Iris-virginica
6.7,3.0,5.2,2.3,Iris-virginica
6.3,2.5,5.0,1.9,Iris-virginica
6.5,3.0,5.2,2.0,Iris-virginica
6.2,3.4,5.4,2.3,Iris-virginica
5.9,3.0,5.1,1.8,Iris-virginica

}

Discuss how your organization (office management for a non-profit) could use an operations management linear programming application to solve a problem or improve a business process.

5. In the table below are simulated data of net photosynthesis rate for five replicate leaves from two plant species, sawgrass (Cladium jamaiscence) and willow (Salix caroliniana). Using the data below and the ‘Analyzing Ecology’ box in Chapter 2 of the text (pg 42), calculate the mean (x), variance (s¬2), standard deviation (s), and standard error (SE) for the two pH conditions. Construct a graph using the means and standard deviations to visually compare photosynthetic capacity between the two species. Make sure to label your axes appropriately and indicate which is the dependent and the independent variable.

Table 1. Photosynthesis rate (µmol CO₂ / m² / sec) for sawgrass and willow leaves.

.

2. Non-Local Strings

For this question, a block is a sequence of 20 characters, where each character is one of the 26 lowercase letters a-z. For example, these are blocks:

iwpiybhunrplsovrowyt
rpulxfsqrixjhrtjmcrr
fxfpwdhwgxtdaqtmxmlf

1. How many different blocks are there?
2. A block is squarefree if no character appears two times consecutively. The first and third example above are squarefree, but the second example is not because of the two consecutive occurrences of r. How many squarefree blocks are there?
3. A block is non-local if the number of characters between any two occurrences of the same character is at least 2. The first example above is non-local. The second example is not, because there are two occurrences of r with no characters between them. The third example is not because there are two occurrences of m with only one character between them. How many non-local blocks are there?
4. A block is k-non-local if the number of characters between any two occurrences of the same character is at least k. Write a formula for the number of k-non-local blocks that is valid for any k∈{0,…,20}.

Sanity check: The formula you get for 2.4 gives the answer to 2.2 when k=1 and gives the answer to 2.3 when k=2.

Jarrid Medical, Inc., is developing a compact machine for kidney dialysis, but the company's chief engineer, Mike Crowe, has trouble controlling the variability of how quickly the fluid moves through the device Medical employers require that the flow per hour be 4.25 liters, 98% of the time. Mr. Crowe, tests the prototype with the following results for each hourly flow test:

With the information obtained from the sample, it is necessary to find the confidence interval that reflects the expected hourly flow in 98% of the time of the developed compact machine and determine if it is satisfying the medical standards. Satisfy the prototype of medical patterns? Present all the steps required to build your confidence interval and an analysis that justifies your answer to the established question.

6. The following is from the General Social Survey, 2016. Interpret the following frequency distribution and the mode, median, or mean, for the variable CCTV, which asks if the American government should have the right to keep people under video surveillance in public areas. Statistics

1. What level of measurement is this variable at?
2. What would be the appropriate measure of central tendency?
3. Write an interpretation of the frequency distribution and measure of central tendency you chose:

Write an interpretation of the frequency distribution and measure of central tendency you chose

Question 1

A student organization wants develop a program for students who are going to graduate in the next 18 months.   To get a sense of how many students might be interested in the program, they want an estimate of the proportion of students who are going to graduate in the next 18 months. They take a simple random sample of 89 students. The data they collected can be found in the column labeled “Graduate in the next 18 month”.   “No” means that the student will not graduate in the next 18 months and would not be of interest to the student organization.   “Yes” means that the student will graduate in the next 16 months and would be of interest to the student organization.

1. The vice president of the student organization has asked you to create an estimate of the proportion of students who will graduate in 18 months. You ask her what level of confidence she wants to have in the estimate and she says that she would like to be 90% confident in the proportion estimate. Create the confidence interval using the sample data. Write a sentence or two to communicate the results to her.

Question 2

A student organization wants to get an estimate of the proportion of students who would attend a Tiny Cowboys concert at Al Lang stadium.   They ask you how many students they should include in their sample to get this estimate.

1. They tell you that a similar concert held at another college close by had 24% of all students attend the concert, they would be ok with being 90% confident and want the estimate to be within 6% of the true value. Calculate the sample size. Write a sentence or two to communicate to the student organization how many students that they should sample.

Question 3

The Student Government Association is interested in an estimate of the mean hours students work per week. They take a simple random sample of 89 students. The data they collected can be found in the column labeled “Number of hours worked per week”.

1. Use the sample data to construct a 95% confidence interval for the mean number hours worked. Write one or two sentences to communicate to the Student Government the estimate and the margin of error.

Question 4

An administrator said that he believes more than 25% of all students will graduate in the next 18 months.

Use the data found in the column labeled “Graduate in the next 18 months” to answer the following questions. “No” means that the student will not graduate in the next 18 months.   “Yes” means that the student will graduate in the next 18 months.

Write the claim in symbolic form

1. Write the null and alternative hypothesis in symbolic form
2. State the test statistic
3. State the p-value
4. State your conclusion about the null hypothesis using a 0.05 significance level
5. Write a sentence to communicate to the administrator your conclusion about his claim

Data:

Students- Graduate in 18 mo- number of hours worked per week

1   No   6.3
2   No   30.1
3   No   15.8
4   No   29.1
5   No   22.6
6   No   13.8
7   No   25.4
8   No   13.1
9   No   25.4
10   No   25.3
11   No   20.6
12   No   31.2
13   No   18.1
14   No   38.1
15   No   22
16   No   33.1
17   No   15.7
18   No   33.9
19   No   21.8
20   No   22.5
21   No   27.1
22   No   25.8
23   No   25.9
24   No   17.2
25   No   28.1
26   No   16.8
27   No   24.8
28   No   37
29   No   15.9
30   No   12.6
31   No   34.7
32   No   23.7
33   No   31.8
34   No   17.7
35   No   19
36   No   35
37   No   33
38   No   10.8
39   No   24.5
40   No   26.9
41   No   31.8
42   No   22.2
43   No   21.8
44   No   26.1
45   No   25.7
46   No   31.4
47   No   25.5
48   No   18.1
49   No   31
50   No   19.8
51   No   15.9
52   No   16.8
53   No   25.4
54   No   21.3
55   No   25
56   No   20.2
57   No   4.8
58   No   37.2
59   No   19.4
60   No   15.7
61   No   18.1
62   No   13.3
63   Yes   31.7
64   Yes   19
65   Yes   23.6
66   Yes   28.4
67   Yes   17.1
68   Yes   26.7
69   Yes   7.1
70   Yes   44.7
71   Yes   31.2
72   Yes   32.7
73   Yes   15.9
74   Yes   19.4
75   Yes   25.6
76   Yes   28.9
77   Yes   27.6
78   Yes   18
79   Yes   29.5
80   Yes   23.6
81   Yes   36.5
82   Yes   41.4
83   Yes   19.6
84   Yes   11.3
85   Yes   29
86   Yes   14.4
87   Yes   27
88   Yes   41.4
89   Yes   13.1

True or False:

a.) In a statistical study, the random variable X = 1, if the house is colonial and X = 0 if the house is not colonial, then it can be stated that the random variable is continuous.

b.) For a continuous distribution, P(X ≤ 10) is the same as P(X<10).

c.) For a continuous distribution, the exact probability of any particular value is always zero.

d.) For a binomial probability experiment, with n = 60 and p =.2, it is appropriate to use the normal approximation to the binomial distribution without continuity correction.

e.) All continuous random variables are normally distributed.

f.) In a binomial distribution the random variable X is discrete.

g.) Events that have no sample space outcomes in common and, therefore cannot occur simultaneously are referred to as mutually independent events.

h.) The probability of an event is the product of the probabilities of the sample space outcomes that correspond to the event.

i.) A classical probability measure is a probability assessment that is based on relative frequency.

Motivated by findings from California that school districts with lower student-teacher ratios have higher average test scores, administrators in New York City recently reviewed the relationship between school- level student-teacher ratios and average test scores within their population of elementary schools. Data for fifth-grade test scores (reading and math) from 1,575 elementary schools yield Y ̄ = 631.7 and sY = 17.8

a) Construct a 95% confidence interval for the mean test score in the population (i.e., of schools in NYC).

b) When NYC administrators divided the population into schools with small (i.e., < 20) and large (i.e., ≥ 20) average class sizes, the 555 schools with small classes had a mean test score of 644 with a standard deviation of 11.7, while the 1,020 schools with large classes had a mean test score of 625 with a standard deviation of 21.1. Is there statistically significant evidence that the schools with smaller class sizes have higher average test scores? Explain.

c) Do these results (likely) represent a causal estimate? Why or why not?

1.The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.63 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.2 inches and a standard deviation of 2.54 inches.

a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?

z =

b) What percentage of men are SHORTER than 6 feet 3 inches? Round to nearest tenth of a percent.

%

c) If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?

z =

d) What percentage of women are TALLER than 5 feet 11 inches? Round to nearest tenth of a percent.

%

2.A population of values has a normal distribution with μ=32.1μ=32.1 and σ=61.9σ=61.9. You intend to draw a random sample of size n=48n=48.

What is the mean of the distribution of sample means?
μ¯x=μx¯=

What is the standard deviation of the distribution of sample means?
(Report answer accurate to 2 decimal places.)
σ¯x=σx¯=

3. A population of values has a normal distribution with μ=114.5μ=114.5 and σ=47σ=47. You intend to draw a random sample of size n=147n=147.

What is the mean of the distribution of sample means?
μ¯x=μx¯=

4. What is the standard deviation of the distribution of sample means (i.e. the standard error)?
(Report answer accurate to 2 decimal places.)
σ¯x=σx¯=

5. SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You are interested in estimating the average SAT score of first year students at your college. If you would like to limit the margin of error of your 95% confidence interval to 25 points, how many students should you sample?

Many consumers pay careful attention to stated nutritional contents on packaged foods when making purchases, so it is important that the information on packages be accurate. The distribution of calorie content has been shown to be approximately normally distributed. A random sample of 12 frozen dinners of a certain type was selected and the calorie content of each one was determined to be 255, 244, 239, 242, 265, 245, 259, 248, 225, 226, 251, and 232. (a) Determine the sample mean and sample standard deviation for these 12 randomly selected frozen dinners. (b) The stated mean calorie content on the box is 240. Construct an appropriate test to see if the actual mean content differs from the stated value at an α = 0.05 significance level. Do not use the T-test feature of your calculator.

Suppose We put five different dice into a hat. The dice have the following number of side:4,6,8,12,20. When we choose a die from the hat, each of the five of the dice are equally likely to appear.

a) What is the probabilty that a “6” appears?

b) Now, suppose a “6” appears, what is the probability is was the 6-sided die that was chosen?

In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and the United Kingdom. They found that a substantially greater percentage of U.K. ads use humor.

(a) Suppose that a random sample of 387 television ads in the United Kingdom reveals that 130 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.K. television ads that use humor. (Round your answers to 3 decimal places.)   

(b) Suppose a random sample of 534 television ads in the United States reveals that 114 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.S. television ads that use humor. (Round your answers to 3 decimal places.)

(c) Do the confidence intervals you computed in parts a and b suggest that a greater percentage of U.K. ads use humor?

(Click to select)NoYes , the U.K. 95 percent confidence interval is (Click to select)abovenot above the maximum value
in the confidence interval for the U.S.