Using Probability And Statistical Analysis To Solve Problems

Weekly attendance sold Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper	Number of chocolate bars
472	6916
413 Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper	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%.