Question

Please i need final report on this program. thanks

% Preparation of Data

tempApril1969 = [36.5 51.5 41.5 47.5 56 50.5 54.5 55 51 61 54 48 50 54 57.5 57.5 64 57 50.5 47.5 47 50.5 52 48 56 65.5 63.5 72 61 54.5]; % temperature data of April, 1969 (shape = 1x30)

tempApril2019 = [39.5 43.5 54 51.5 42 52.5 55 63 46.5 51 46 55.5 66.5 63 53.5 54 54 56.5 65.5 58.5 57.5 54.5 64.5 61 56 55.5 52.5 49 51.5 52.5]; % temperature data of April, 2019 (shape = 1x30)

days = 1:30; % days of month April

tempDiff = tempApril2019 - tempApril1969; % Taking difference of both datas

figure(1) % start plotting 1st figure

title('April 2019 vs. April 1969'); % setting title for the figure

bar(days,[tempApril1969',tempApril2019']); % Creating a bar chart of 2 datas simultaneously w.r.t. days in april

hold on % retain current plot when adding new plots

legend('1969','2019') % Setting legends to represent different colors of bars

xlabel('Day in April'); % Label X-axis

ylabel('Temperature') % Label Y-axis

figure(2) % start plotting second figure

plot(days,tempDiff, '--*'); % plot a line graph of temperature difference vs. days

title('Difference in Temperatures'); % setting title for the graph

xlabel('Days'); % Label X-axis

ylabel('Temperature'); % Label Y-axis

%% Calculate mean

N = length(days); % calculating length of days vector for further use

mean2019 = sum(tempApril2019)/N; % Calculating mean of temperatures of April, 2019

mean1969 = sum(tempApril1969)/N; % Calculating mean of temperatures of April, 1969

difference_of_mean = mean2019 - mean1969; % Taking difference of mean

%% Calculate standard deviation

% X² = X * transpose(X)

% Variance(X) = ( ( X - Mean(X) )² / No. of elements in X )

% Standard Deviation = Square root (Variance)

variance2019 = ((tempApril2019-mean2019)*(tempApril2019-mean2019)')/N; % Calculating variance of the temperature in April, 2019

variance1969 = ((tempApril1969-mean1969)*(tempApril1969-mean1969)')/N; % Calculating variance of the temperature in April, 1969

sd2019 = sqrt(variance2019); % Taking square root of variance to get standard deviation of temperature in April, 2019

sd1969 = sqrt(variance1969); % Taking square root of variance to get standard deviation of temperature in April, 1969

%% Correlation Coefficient

covariance = (tempApril2019-mean2019)*(tempApril1969-mean1969)'; % calculating covariance of temperatures in April, 1969 and April, 2019

correlationCoeff = covariance/(sd2019*sd1969); % Calculating correlation coefficient i.e. covariance_1_2 / (standard_deviation_1 * standard_deviation_2)

%% Display

% displaying the message where %f is replaced by the float value of the

% corresponding variable in the list input after the closing of quotes

msg = sprintf("The mean of temperature in 1969: %f\nThe mean of temparature in 2019: %f\nDifference of mean: %f\nStandard deviation in 1969: %f\nStandard deviation in 2019: %f\nCorrelation Coefficient: %f",mean1969,mean2019,difference_of_mean,sd1969,sd2019,correlationCoeff);

disp(msg)

Answer 1

answer:

% Preparation of Data

tempApril1969 = [36.5 51.5 41.5 47.5 56 50.5 54.5 55 51 61 54 48 50 54 57.5 57.5 64 57 50.5 47.5 47 50.5 52 48 56 65.5 63.5 72 61 54.5]; % temperature data of April, 1969 (shape = 1x30)

tempApril2019 = [39.5 43.5 54 51.5 42 52.5 55 63 46.5 51 46 55.5 66.5 63 53.5 54 54 56.5 65.5 58.5 57.5 54.5 64.5 61 56 55.5 52.5 49 51.5 52.5]; % temperature data of April, 2019 (shape = 1x30)

days = 1:30; % days of month April

tempDiff = tempApril2019 - tempApril1969; % Taking difference of both datas

figure(1) % start plotting 1st figure

title('April 2019 vs. April 1969'); % setting title for the figure

bar(days,[tempApril1969',tempApril2019']); % Creating a bar chart of 2 datas simultaneously w.r.t. days in april

hold on % retain current plot when adding new plots

legend('1969','2019') % Setting legends to represent different colors of bars

xlabel('Day in April'); % Label X-axis

ylabel('Temperature') % Label Y-axis

figure(2) % start plotting second figure

plot(days,tempDiff, '--*'); % plot a line graph of temperature difference vs. days

title('Difference in Temperatures'); % setting title for the graph

xlabel('Days'); % Label X-axis

ylabel('Temperature'); % Label Y-axis

%% Calculate mean

N = length(days); % calculating length of days vector for further use

mean2019 = sum(tempApril2019)/N; % Calculating mean of temperatures of April, 2019

mean1969 = sum(tempApril1969)/N; % Calculating mean of temperatures of April, 1969

difference_of_mean = mean2019 - mean1969; % Taking difference of mean

%% Calculate standard deviation

% X2 = X * transpose(X)

% Variance(X) = ( ( X - Mean(X) )2 / No. of elements in X )

% Standard Deviation = Square root (Variance)

variance2019 = ((tempApril2019-mean2019)*(tempApril2019-mean2019)')/N; % Calculating variance of the temperature in April, 2019

variance1969 = ((tempApril1969-mean1969)*(tempApril1969-mean1969)')/N; % Calculating variance of the temperature in April, 1969

sd2019 = sqrt(variance2019); % Taking square root of variance to get standard deviation of temperature in April, 2019

sd1969 = sqrt(variance1969); % Taking square root of variance to get standard deviation of temperature in April, 1969

%% Correlation Coefficient

covariance = (tempApril2019-mean2019)*(tempApril1969-mean1969)'; % calculating covariance of temperatures in April, 1969 and April, 2019

correlationCoeff = covariance/(sd2019*sd1969); % Calculating correlation coefficient i.e. covariance_1_2 / (standard_deviation_1 * standard_deviation_2)

%% Display

% displaying the message where %f is replaced by the float value of the

% corresponding variable in the list input after the closing of quotes

msg = sprintf("The mean of temperature in 1969: %f\nThe mean of temparature in 2019: %f\nDifference of mean: %f\nStandard deviation in 1969: %f\nStandard deviation in 2019: %f\nCorrelation Coefficient: %f",mean1969,mean2019,difference_of_mean,sd1969,sd2019,correlationCoeff);

disp(msg)

Sample Run:

The mean of temperature in 1969:53.833333 The mean of tenparature in 2019: 54.20806 Difference of mean: .356667 Standard deviation in 1969: 7.133645 Standard deviation in 2019: 6.576995 Correlation Coefficient: -0,506673 Figure 1 80 1969 2019 70 60 50 40 30 20 10 15 10 20 25 30 Day in April

Figure 2 Difference in Temperatures 20 15 10 -10 -15 20 30 25 20 15 10