# Using Probability And Statistical Analysis To Solve Problems

• December 28, 2023/
 Weekly attendance sold Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free 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. 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%.