Question

All pages please

15 year wolf is a MATLAB: 2. ao pes/2 pes each) Translating sim rd P in later animals bom in each array with 50 rows and 15 the use it number span Once you define a variable u are free to total of a) Define a MATLAB variable wolvesbornbys that the contains b) Define a MATLAB variable the difference in bom from year to year by state. wolfbirthchange animals that contains c) Define a MATLAB variable wolvesbornrx that contains the number of animals bom in Texas (row39) d efine a MATLAB variable wolvesbornsince2010 that contain the number of animals born in the us since 2010. e) Define a MATLAB variable wolfTxpercentage that contains the percentage of animals born in Te ompared to all animals born in the US. Final Exa Spring 2017

Answer 1

Q2)
a. wolvesbornbystate=sum(wolf,2)
This will give the sum of each row.

b. wolfbirthchange=diff(wolf,[],2)
This will calculate the difference between two elements of adjacent columns, row-wise. ie- difference between columns' data.

c. wolvesbornTX=sum(wolf(39,:))
This will give the sum of all elements in row 39.

d. wolvesbornsince2010=sum(sum(wolf(1:50,10:15)))
wolf(1:50,10:15) will return all elements in columns 10 to 15 (years 2010 to 2015), 1:50 is to select all rows in those columns.
sum(wolf(1:50,10:15)) will calculate sums of each of these returned columns
sum(sum(wolf(1:50,10:15))) will finally give the sum of all these above sums.

e. wolfTxPercentage=wolvesbornTX/(sum(sum(wolf))- wolvesbornTX) *100
sum(sum(wolf)) will give the sum of all the elements in the matrix and (sum(sum(wolf))- wolvesbornTX) will give the sum of all elements except Texas.
You may replace wolfTxPercentage with sum(wolf(39,:)) if required.

Q3)
a. 18 students from Section 1 averaged around 8.5 hours of sleep a day.
b. For this we need to add up all the uppper edge values of the each bar of the histogram. The first bar is approximately 8, the second, approximately 13, the third, 18 and so on.
So, the approximate number of students in section 1 shown here is-8+13+18+2+3+1+1=46
c. No of students in section 3 sleeping an average of 8 hours=20
Total No of students in section 3=58
Therefore, percentage of students in section 3 sleeping a average of 8 hours =20/58*100=34.48%
d. Both distributions are similar in that both have their highest values around the same points,ie- most students in both sections sleep around 8 hours.
Data for section 3 is slightly more continuous.

Q4)
a. i. Median=45; this is the line inside the box.
ii. The Inter Quartile Range=0.41 to 0.45; this is 25 to 75 percentile: the area spanned by the box.
iii. The 75th percentile is 0.45; the upper edge of the box.
iv. The largest value is 0.52, the upper end of the upper whisker.
v. Top fence value- not entirely sure what this means; do you require the upper inner fence or the upper outer fence (outlier) or something else?

b. The position of the top whisker represents the highest percentage of students reporting caffeine.

c. The notches represent the confidence interval around the median; it is the range within which we can say with fair certainty that the true value lies.

d. 1. Caffeince consumption is greater among males than females.
2. The value of the lowest percentage of the males consuming caffeine is around the value of the median for females consuming caffeine.
3. The spread of the percentage of females reporting caffeine is greater than that of the males.