Statistics on the Number of Students Who Play Cricket, Hockey, and Volleyball

1. For ease with the calculations let’s use letters to represent the games.

That is  Cricket

            H for Hockey         V for Volley

1. Drawing the Venn diagram to represent the scenario

The number of students who play:

Cricket 50

Hockey 50

Volley 40

Cricket and Hockey 5

Hockey and Volley 10

Cricket and Volley 5

1. If every student play at least one game this means either a student play one, two or three games but there is no single student who does not participate in games.

The number of students will therefore be  

This gives the total number of students to be 100

1. The number of students who play cricket only

Total playing cricket is 50, all three games are played by 10 students.

Cricket and hockey 5 students and finally cricket and volley 5 students.

Hence cricket only will be.  

• The number of students who are playing hockey and volleyball only, but no cricket are 10 students.

This number can be obtained directly from the Venn diagram by checking the intersection of H and V.

1. The survey involves 170 respondents on their interests in Astro Channels. Using parameters

Astro prima be P

Astro Ria be R

Astro Mustika be M   

P and M    

Total 170 respondents

1. Assuming no respondent like all the 3 channels that means the value of x is zero. Then the number of respondents who like Astro Musika only will be

Total who like Astro Musika

Then

And

M only will therefore be

1. The number of respondents who like at least 2 channels.    

Total will be

• Respondents who like , this value can be observed directly from the Venn diagram.

Question 2

2. The frequency tables

• Value of the median

The median from the frequency distribution table will be obtained using the formulaThe parameters used areL the = lower boundaryTotal frequency cumulative frequency above the box frequency in the box class interval sizemedian is the middle hence The 25 will be a rough idea (box location) of the median.Hence from the cumulative frequency column we pick a number that is the first one to be greater than 25. In our case it will be 33.From here we draw a box in the row containing this number. This box assist obtains the median.Replacing this values in the above formula will give the median as Obtaining the Standard DeviationFrom the frequency distribution we obtain the standard deviation using the formulaThe values needed in the formula can be obtained directly from the frequency table above.Hence inserting the values in the formula, we have the Standard deviation asThe value will be

• Drawing the ogive graph

The table will be          

1. The mean of the data is 75.

Mean is calculated using the formula Simplifying this equation gives

1. The mode is the marks with the highest frequency. If the mode of the data is 88 then the value of x should be at a minimum 11.       

Question 3

1. Probability a component is made by machine A or B        

1. The probability that all the two components are made by machine B        

ways on which the components can be arranged will have

 as the definitive answer.

• The probability of A and B and C

This gives 0.035

Since there are nine ways of arranging the products depending on which machine produced them then the answer is raised to power 3 to give       

1. The probability of not acceptable 

1. The probability of B given not acceptable 

1. Probability of acceptable        

Question 4

143.25

1. Sketching the graphs 

1. Integrating with respect to x

The equation

When integrated with respect to x gives

On the other hand, when integrated with respect to x gives 1. With thus the area under the two curves when obtained using integration with respect to x will be 1Integrating with respect to y 

Question 5

1. You given and Then Is On the other hand Therefore, the value of  which the value of  as it is positive hence the absolute value will be the same.    

The is to verify that To begin we computer This means Thereafter we computer This equals (AB) which is  Since the final solution of (AB)C  then we have successfully proven that

1. To initiate the solution lets first replace the items bought by a set of letters

Sheets of crats papers be S

Boxes of markers be B

Glue sticks be G

1. For Johny For Sara nd for Chong 

1. Using Cramer’s rule to solve the system of linear equations

The Matrix computed is          

 From here we write down the main matrix. This is We then find the determinant of this matrix. Which will be Then the 1^st column of the main matrix is replaced by the solution vector and the determinant of the resultant matrix obtained The determinant will be given by e next step will be to replace the 2^nd column of the main matrix with the solution vector and obtain the determinant The resultant matrix will be The determinant will be Thereafter we replace the 3^rd column of the matrix with the solution vector and determine the resultant determinant. 

The matrix is The determinant will be  Now the value of the items will be obtained by  From the solution obtained the unit costs will be as follows

For Craft paper Box of Markers Glue sticks 

References

Freedman, D. (2005). Statistical Models: Theory and Practice. Cambridge University Press.

Gut, A. (2005). Probability: A Graduate Course. Springer-Verlag.

Katz, V. J. (2008). A history of mathematics . Boston: Addison-Wesley.

Merriam-Webster. (2017). Integral Calculus – Definition of Integral calculus.

Ruskey, F., Savage, C. D., & Wagon, S. (2006). The Search for Simple Symmetric Venn Diagrams. Notices of the AMS, 1304–11.

Sandifer, E. (2003). How Euler Did It. The Mathematical Association of America.