Using Probability And Statistical Analysis To Solve Problems

Weekly attendance sold

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Number of chocolate bars

472

6916

413

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5884

503

7223

612

8158

399

6014

538

7209

455

6214

Population and Sample Data

The above is a population. The above is a population since a population entails all the members in a group. The data above covers 7 weeks. A sample is a representative of a population thus for the data above to be a sample it would have to be a part of data from 2 months or more.

Standard deviation

Weekly attendance sold (x)

(x- x?)

(x- x?)^2

472

-12.57

158.04

413

-71.57

5122.47

503

18.43

339.61

612

127.43

16238.04

399

-85.57

7322.47

538

53.43

2854.61

455

-29.57

874.47

Sum

32909.71

Standard deviation

68.57

Mean = (472+413+503+612+399+538+455) / 7 = 484.57

Standard deviation = 68.57

Inter Quartile Range

5884, 6014, 6214, 6916, 7209, 7223, 8158

IQR = Q3 – Q2

Q3 = ¾(n+1)th term

Q3 = ¾ * (7+1) = 6 = 7223

Q1 = ¼ (n+1)th term

Q1 = ¼ (7+1) = 2 = 6014

IQR = 7223 – 6014= 1209

IQR is better than standard deviation since it best describes the spread in an empirical statistical distribution of data. The IQR shows that the variation of chocolate bars is 1,209 in the 7 weeks. 

  1. Correlation coefficient

 (x)

 (y)

(xy)

(x^2)

(y^2)

            472

           6,916

           3,264,352

             222,784

           47,831,056

            413

           5,884

           2,430,092

             170,569

           34,621,456

            503

           7,223

           3,633,169

             253,009

           52,171,729

            612

           8,158

           4,992,696

             374,544

           66,552,964

            399

           6,014

           2,399,586

             159,201

           36,168,196

            538

           7,209

           3,878,442

             289,444

           51,969,681

            455

           6,214

           2,827,370

             207,025

           38,613,796

Totals

         3,392

         47,618

         23,425,707

         1,676,576

         327,928,878

= ((7*23,425,707)-3,392*47,618)/[(7*1,676,576-(3,392^2))*(7*4327,928,878 – (47,618^2))]

= 0.97

The correlation shows that there is a high positive relationship between weekly attendance of students and the number of chocolate bars. Thus, this information can be used to drive up the sales of chocolate bars by increasing the number of weekly attendance by the students. 

Weekly attendance sold

Number of chocolate bars

472

6916

413

5884

503

7223

612

8158

399

6014

538

7209

455

6214

Regression equation

x

y

x2

xy

472

6916

222784

3264352

413

5884

170569

2430092

503

7223

253009

3633169

612

8158

374544

4992696

399

6014

159201

2399586

538

7209

289444

3878442

455

6214

207025

2827370

Sum

3392

47618

1676576

23425707

Average

484.57

6802.57

The independent variable is weekly attendance while the dependent variable is the number of chocolate bars. Thus, weekly attendance affects the number of chocolate bars sold.

Thereby;

1 =(∑XY – (∑X∑Y)/7)/( ∑X2 – ((∑X)2)/n))

1 = (23425707 – (3392*47618)7) / (1676576 – (33922)/7)

1 = 10.7 

0 = y? – 1  x?

0 = 6802.57 – 10.7 * 484.57

0 = 1628.69 

Thus, y = 1628.69 + 10.7x

Therefore, a unit increase in weekly attendance increases the number of chocolate bars sold by 10.7 units. Thus, when Holmes close, the number of chocolate bars sold remains constant at 1,628 units.

When 10 extra students show up, the number of chocolate bars sold increases by 107 units. Thus, the total number of chocolate bars sold will be 1735 bars.

  1. Coefficient of determination

r = 1 )2*(∑X2 – ((∑X)2/n))/( ∑Y2 – (∑Y)2)n)

r = (10.72*(1,676,576 – ((3,392)2/7)) / (327,928,878 – ((47,6182)/7))

r = 0.9

From the coefficient of determination, it can be concluded that 90% of the variation are explained by factors in the model while 10% can be explained by factors not in the model. 

Scientific training

Grassroots training

Total

Recruited from Holmes students

35

92

127

External recruitment

54

12

66

Total

89

104

193

Player from Holmes OR receiving Grassroots training

= (127/193) + (104/193) – (92/193)

= 0.72

  1. External and scientific training

= 54/193 = 0.28

  1. Player from Holmes, probability that he is in scientific training?

= (127/193)* (35/89)

= 0.26

  1. Is training independent from recruitment?

Picking a player from scientific training = 89/193 = 0.46

Picking a player from external recruitment = 66/193 = 0.34

Since the two probabilities are different, then training and recruitment are dependent.

1 in 10 purchases thus 1/10 probability

P(X = 0) + P(X = 1) + P(X =2)

= [ [

= *0.43 +  *0.05 +  *0.005

=1*0.43 + 8*0.05 + 28*0.005

= 0.97

P(X = 9 | λ = 4) =  

 = (262144 * 0.018316) / 362880

= 0.0132 

Current price – 1.1 million

Months- 12 months

Std deviation – 385000

  1. Probability to sell over 2 million

P(z > 2 million)

P(((x-μ)/σ) < ((2,000,000 – 1,100,000)/385,000)) 

Standard Deviation and Mean

= P(0< z < 2.34) = 0.4904

P (z > 2.34)

= P(z > 2.34) = 0.5 – P(0 <z < 2.34)

= 0.5 – 0.4904

= 0.0096

 The probability to sell over 2 million is 0.96%

  1. Apartment will sell for over 1 million but less than 1.1 million

P(1 < z < 1.1) = P(0 < (( x – μ) / σ) < (1.1 – 1) / 0.385)

 = P (0 < z < 0.26)

= 0.1026

Probability to sell an apartment for over 1 million but less than 1.1 million is 10.26%

We can use the z-distribution to test the assistant’s research findings against mine since the distribution is not normal. According to Townsend (2002), data does not have to be normal for a z-test. However, the variance should be approximately equal. 

p? =sample proportion =11/45

p = population proportion = 0.3

n = sample size

z = (p? – p)/()

z = (0.24 – 0.3)/((0.3*0.7)/45))

z = -0.88

z = -0.88 has a probability of 0.189

Thus, the probability of 30% of the investors to be willing to commit $1 million or more to the fund is 18.9%.

Weekly attendance sold

Number of chocolate bars

472

6916

413

5884

503

7223

612

8158

399

6014

538

7209

455

6214

The above is a population. The above is a population since a population entails all the members in a group. The data above covers 7 weeks. A sample is a representative of a population thus for the data above to be a sample it would have to be a part of data from 2 months or more.

  1. Standard deviation

Weekly attendance sold (x)

(x- x?)

(x- x?)^2

472

-12.57

158.04

413

-71.57

5122.47

503

18.43

339.61

612

127.43

16238.04

399

-85.57

7322.47

538

53.43

2854.61

455

-29.57

874.47

Sum

32909.71

Standard deviation

68.57

Mean = (472+413+503+612+399+538+455) / 7 = 484.57

Standard deviation = 68.57

Inter Quartile Range

5884, 6014, 6214, 6916, 7209, 7223, 8158

IQR = Q3 – Q2

Q3 = ¾(n+1)th term

Q3 = ¾ * (7+1) = 6 = 7223

Q1 = ¼ (n+1)th term

Q1 = ¼ (7+1) = 2 = 6014

IQR = 7223 – 6014= 1209

IQR is better than standard deviation since it best describes the spread in an empirical statistical distribution of data. The IQR shows that the variation of chocolate bars is 1,209 in the 7 weeks.

Correlation coefficient

 (x)

 (y)

(xy)

(x^2)

(y^2)

            472

           6,916

           3,264,352

             222,784

           47,831,056

            413

           5,884

           2,430,092

             170,569

           34,621,456

            503

           7,223

           3,633,169

             253,009

           52,171,729

            612

           8,158

           4,992,696

             374,544

           66,552,964

            399

           6,014

           2,399,586

             159,201

           36,168,196

            538

           7,209

           3,878,442

             289,444

           51,969,681

            455

           6,214

           2,827,370

             207,025

           38,613,796

Totals

         3,392

         47,618

         23,425,707

         1,676,576

         327,928,878

= ((7*23,425,707)-3,392*47,618)/[(7*1,676,576-(3,392^2))*(7*4327,928,878 – (47,618^2))]

= 0.97

The correlation shows that there is a high positive relationship between weekly attendance of students and the number of chocolate bars. Thus, this information can be used to drive up the sales of chocolate bars by increasing the number of weekly attendance by the students.

Weekly attendance sold

Number of chocolate bars

472

6916

413

5884

503

7223

612

8158

399

6014

538

7209

455

6214

Regression equation

x

y

x2

xy

472

6916

222784

3264352

413

5884

170569

2430092

503

7223

253009

3633169

612

8158

374544

4992696

399

6014

159201

2399586

538

7209

289444

3878442

455

6214

207025

2827370

Sum

3392

47618

1676576

23425707

Average

484.57

6802.57

The independent variable is weekly attendance while the dependent variable is the number of chocolate bars. Thus, weekly attendance affects the number of chocolate bars sold.

Thereby;

1 =(∑XY – (∑X∑Y)/7)/( ∑X2 – ((∑X)2)/n))

1 = (23425707 – (3392*47618)7) / (1676576 – (33922)/7)

1 = 10.7 

0 = y? – 1  x?

0 = 6802.57 – 10.7 * 484.57

0 = 1628.69 

Thus, y = 1628.69 + 10.7x

Therefore, a unit increase in weekly attendance increases the number of chocolate bars sold by 10.7 units. Thus, when Holmes close, the number of chocolate bars sold remains constant at 1,628 units.

When 10 extra students show up, the number of chocolate bars sold increases by 107 units. Thus, the total number of chocolate bars sold will be 1735 bars. 

  1. Coefficient of determination

r = 1 )2*(∑X2 – ((∑X)2/n))/( ∑Y2 – (∑Y)2)n)

r = (10.72*(1,676,576 – ((3,392)2/7)) / (327,928,878 – ((47,6182)/7))

r = 0.9

From the coefficient of determination, it can be concluded that 90% of the variation are explained by factors in the model while 10% can be explained by factors not in the model.

Scientific training

Grassroots training

Total

Recruited from Holmes students

35

92

127

External recruitment

54

12

66

Total

89

104

193

Player from Holmes OR receiving Grassroots training

= (127/193) + (104/193) – (92/193)

= 0.72

  1. External and scientific training

= 54/193 = 0.28

  1. Player from Holmes, probability that he is in scientific training?

= (127/193)* (35/89)

= 0.26

  1. Is training independent from recruitment?

Picking a player from scientific training = 89/193 = 0.46

Picking a player from external recruitment = 66/193 = 0.34

Since the two probabilities are different, then training and recruitment are dependent.

1 in 10 purchases thus 1/10 probability

P(X = 0) + P(X = 1) + P(X =2)

= [ [

= *0.43 +  *0.05 +  *0.005

=1*0.43 + 8*0.05 + 28*0.005

= 0.97

P(X = 9 | λ = 4) =  

 = (262144 * 0.018316) / 362880

= 0.0132

Current price – 1.1 million

Months- 12 months

Std deviation – 385000

  1. Probability to sell over 2 million

P(z > 2 million)

P(((x-μ)/σ) < ((2,000,000 – 1,100,000)/385,000)) 

= P(0< z < 2.34) = 0.4904 

P (z > 2.34)

= P(z > 2.34) = 0.5 – P(0 <z < 2.34)

= 0.5 – 0.4904

= 0.0096

 The probability to sell over 2 million is 0.96%

  1. Apartment will sell for over 1 million but less than 1.1 million

P(1 < z < 1.1) = P(0 < (( x – μ) / σ) < (1.1 – 1) / 0.385)

 = P (0 < z < 0.26)

= 0.1026

Probability to sell an apartment for over 1 million but less than 1.1 million is 10.26%

We can use the z-distribution to test the assistant’s research findings against mine since the distribution is not normal. According to Townsend (2002), data does not have to be normal for a z-test. However, the variance should be approximately equal. 

p? =sample proportion =11/45

p = population proportion = 0.3

n = sample size 

z = (p? – p)/()

z = (0.24 – 0.3)/((0.3*0.7)/45))

z = -0.88

z = -0.88 has a probability of 0.189

Thus, the probability of 30% of the investors to be willing to commit $1 million or more to the fund is 18.9%.