Hi I am doing a statistics course for my CGA (Certified General Accountant) and I am having trouble on a assignment and I was hoping that some one could over look my work and let me know what you think. This assignment is a lot of algebra and statistics math equations. If any one could be help it would be sincerely appreciated.
Here is the assignment :
Prepare this assignment in a format that is compatible with Microsoft Office. If you are unsure about whether your files are compatible, check with your instructor or contact the Student Support Centre.
Assignment 1 covers the material you studied in Lessons 1 to 5 and is worth 20 percent of your final grade for this course. It is made up of nine problems with multiple parts. The distribution of marks is indicated in parentheses next to each question.
Important! The solutions to assignment problems must show all important steps such as formulas and procedures used in the solutions (or references to formulas—e.g., textbook formula 10.7—when formulas are difficult to print). Full credit will not be given if only the final answers are displayed. On the other hand, partial marks will be awarded for correct formulas and procedures even if final answers contain calculation errors.
1. A sociologist interested in youth culture is conducting a study on Twitter. To determine how many times students log in to their accounts, she obtains a sample of 10 students. Students are asked how many times they log in to their account per month. The responses are shown below. (4 marks)
1 140 2 174 3 122 4 250 5 145 6 120 7 180 8 212 9 142 10 125
a. According to the sample the sociologist collected, how many times on average did a student access his/her Twitter account each month?
b. Calculate and interpret the inter-quartile range.
c. Calculate and interpret the 80th percentiles.
d. Calculate the z-scores and state whether or not there are any outliers in this data set. Explain
2. To compare the users of Twitter to those of Facebook, the sociologist asked another 10 students to indicate the number of times they log in to their Facebook account each month. The average number of times a student accessed his/her account per month was 150 with standard deviation of 25. (4 marks)
a. Which of the two groups (users of Twitter vs. Facebook) is more dispersed relative to mean in their usage and why?
3. Investor ABC has two stocks: A and B. Each stock may increase in value, decrease in value, or remain unchanged. Consider the experiment of investing in the two stocks and observing the change (if any) in value. (4 marks)
a. How many experimental outcomes are possible?
b. Show a tree diagram for the experiment.
c. How many of the experimental outcomes result in an increase in value for at least one of the two stocks?
d. How many of the experimental outcomes result in a decrease in value for both stocks?
4. The Wildlife Research Institute conducted a survey to learn about whether or not people in certain regions of the country support wildlife. The data collected by the Institute is shown in the table below. (8 marks)
Support Wildlife Region Yes No 1 125 50 2 160 55 3 280 75 4 220 35 Total 785 215
a. What is the probability of finding a person who is from Region 1 and does not support wildlife?
b. If a person is from Region 2, what is the probability that he/she supports wildlife?
c. If a particular person supports wildlife, what is the probability that he/she is from Region 4?
d. If a person is randomly selected from the four regions above, what is the probability that he/she supports wildlife?
5. A survey of 254 landscaping companies showed that 155 of them advertised in newspaper A, 152 advertised in newspaper B, and 110 advertised in both.
a. What is the probability that a landscaping company advertised in at least one of the two news papers?
b. What is the probability that a landscaping company did not advertise in either news paper? (4 marks)
6. To investigate how often families eat out, The Surveying Company surveyed 500 adults living with children under the age of 18. The survey results are shown in the table below. (6 marks)
Number of Meals at a Restaurant Per Week Number of Survey Responses 0 14 1 12 2 30 3 34 4 36 5 120 6 114 7 or more 140
For a randomly selected family with children under the age of 18, compute the probability that
a. the family eats at a restaurant at least three meals during the week.
b. the family eats at a restaurant two or fewer meals during the week.
c. the family eats no meals at a restaurant during the week.
7. A CPA student has to take ten courses to graduate. If none of the courses are prerequisites for others, how many groups of three courses can the student select
for the coming semester? (2 marks)
8. Thirty percent of the workers in company XYZ take public transportation daily to go to work. In a sample of ten workers, what is the probability that (6 marks)
a. four workers take public transportation to work daily?
b. at most, two people take public transportation to work daily?
c. at least three people take public transportation to work daily?
9. Customers arrive at a particular ATM at the rate of 30 customers per hour.
a. What is the probability of three customers arriving in a five minute-time interval?
b. What is the probability of at least three customers arriving in a five-minute interval?
c. What is the probability of no customers arriving in a five-minute interval?
Ok and here is what I have done so far:
1a) The average time the average student accesses their twitter account is 161 times a month.
120+122+125+140+142+145+174+180+212+250/10 = 161
b) The inter quartile range for the twitter usage is 87 this means that the difference between
the 1st quartile and the 3rd quartile is 87.
IQR = Q3 - Q1 Q1 = 25/100*10 = 2.50 between 122 and125 rounded to 125
Q3 = 75/100* 10 = 7.50 between 174 and 180 rounded to 180
IQR = 180 – 125 = 87
c) The 80th percentile is 215 this means that 80 percent of this study Twitter users are logging into their accounts less than 215 times per month.
80/100*10 + ½ = 8.5 this falls between 180 and 215 therefore rounded to 215
d) The z-scores for student Twitter usage are measuring how far the specified student’s Twitter usage is from the mean or average, the average usage being 161 times a month. The z-scores for each student are as follows:
zi = xi – ͞x ͞x = 161 s= ∑(xi - ͞x)2
s = (120-161)2+ (122-161)2+ (140-161)2+ (145+161)2+ (174-161)2+ (180+161)2+ (212-161)2+ (250-161)2 =16,608
10-1 =√1845 s = 42.95 rounded to 43
Student 1 140 - 161/43 = -.488372
Student 2 174 - 161/43 = .302326
Student 3 122 - 161/43 = -.930233
Student 4 250 - 161/43 = 2.069767
Student 5 145 - 161/43 = -.372093
Student 6 120 - 161/43 = .953488
Student 7 180 - 161/43 = .441860
Student 8 212 -161/43 = 1.186047
Student 9 142 - 161/43 = -.44186
Student 10 125 - 161/43 = -.837209
Z-scores that are more than +3 or less than -3 are outliners, theses are data that do not fit in with the other observations, they are unusually large or unusually small amounts. If outliners are found in z-scores they must be reviewed to find if there was a mistake in the data and needs to be corrected, or it is possible that the outliner data is in fact correct it may stay in the data set. This data sample of Twitter users does not contain any outliners.
2a. The standard deviation is the square root of the variance, it measures the dispersion of data samples from the mean, a high standard deviation indicates that the data samples are spread over a large range and a low standard deviation indicates that the data samples are close to the mean. In the comparison between Twitter and Facebook users the standard deviation for Twitter users is 43 whereas the standard deviation for Facebook users is 25. This comparison indicates that relative to the mean the data samples for Facebook users are closer to the mean than Twitter users therefore Twitter users are more disbursed from the mean, mean being the average we can say that the samples of data of Facebook users are closer to the average than the samples of Twitter users.
3a. The possible outcomes for stocks A and B are to increase in value, to decrease in value or to stay the same, there are 9 possible experimental outcomes.
Possible Experimental Outcomes are as follows: stock A (n1 = 3) has three possible outcomes, stock B (n2 = 3) also has 3 possible outcomes (3)(3) = 9 possible outcomes.
A= Stock A B = Stock B I= increase in value D= decrease in value S= stay the same
Possible Experimental Outcomes:
1 AI BI
2 AI BD
3 AI BS
4 AS BS
5 AS BI
6 AS BD
7 AD BI
8 AD BS
9 AD BD
b. Tree diagram
A= Stock A B = Stock B I= increase in value D= decrease in value S= stay the same
The experimental outcomes (sample points) are as follows:
1 stock A increases in value and stock B increases in value
2 stock A increases in value and stock B decreases in value
3 stock A increases in value and stock B stays the same (remains unchanged)
4 stock A decreases in value and stock B decreases in value
5 stock A decreases in value and stock B increases in value
6 stock A decreases in value B stays the same (remains unchanged)
7 stock A stays the same (remains unchanged) and stock B stays the same (remains unchanged)
8 stock A stays the same (remains unchanged) and stock B increases in value
9 stock A stays the same (remains unchanged) and stock B decreases in value
c. There are four experimental outcomes that result in an increase in value for at least one of the
d. There is one experimental outcome that results in a decrease in value for both stocks.
4a. The probability of finding a person who is from region 1 that does not support wildlife is
b. The probability of he or she is a supporter of wildlife is
c. The probability of a person who supports wildlife being from Region 4 is
d. The probability of a finding randomly selected person from any of the four regions that supports wildlife is
5a. The probability that a landscaping company advertized in at least one of the two newspapers is
So far I am stuck on question 5 due to the fact that the survey is of 254 companies and that 155 of them are in newspaper A and 152 are in newspaper B and 110 are in both A & B which leaves me very confused... So to me it seem the survey should be of 307 not 254 or another theory would be:
152+155=307-110=197 254-197=57 did not advertise in the newspaper