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1 year ago

A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp

Mathematics

1 answer:

abruzzese [7]1 year ago

7 0

Step-by-step explanation:

"A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp."

"(a) During a morning activity of the camp, these 12 players have to randomly group into six pairs of two players each."

"(i) Find the total number of possible ways that these six pairs can be formed."

The order doesn't matter (AB is the same as BA), so use combinations.

For the first pair, there are ₁₂C₂ ways to choose 2 people from 12.

For the second pair, there are ₁₀C₂ ways to choose 2 people from 10.

So on and so forth. The total number of combinations is:

₁₂C₂ × ₁₀C₂ × ₈C₂ × ₆C₂ × ₄C₂ × ₂C₂

= 66 × 45 × 28 × 15 × 6 × 1

= 7,484,400

"(ii) Find the probability that each pair contains players of the same gender only. Correct your final answer to 4 decimal places."

We need to find the number of ways that 6 boys can be grouped into 3 pairs. Using the same logic as before:

₆C₂ × ₄C₂ × ₂C₂

= 15 × 6 × 1

= 90

There are 90 ways that 6 boys can be grouped into 3 pairs, which means there's also 90 ways that 6 girls can be grouped into 3 pairs. So the probability is:

90 × 90 / 7,484,400

= 1 / 924

≈ 0.0011

"(b) During an afternoon activity of the camp, 6 players are randomly selected and 6 one-on-one matches with the coach are to be scheduled.

(i) How many different schedules are possible?"

There are ₁₂C₆ ways that 6 players can be selected from 12. From there, each possible schedule has a different order of players, so we need to use permutations.

There are 6 options for the first match. After that, there are 5 options for the second match. Then 4 options for the third match. So on and so forth. So the number of permutations is 6!.

The total number of possible schedules is:

₁₂C₆ × 6!

= 924 × 720

= 665,280

"(ii) Find the probability that the number of selected male players is higher than that of female players given that at most 4 females were selected. Correct your final answer to 4 decimal places."

If at most 4 girls are selected, that means there's either 0, 1, 2, 3, or 4 girls.

If 0 girls are selected, the number of combinations is:

₆C₆ × ₆C₀ = 1 × 1 = 1

If 1 girl is selected, the number of combinations is:

₆C₅ × ₆C₁ = 6 × 6 = 36

If 2 girls are selected, the number of combinations is:

₆C₄ × ₆C₂ = 15 × 15 = 225

If 3 girls are selected, the number of combinations is:

₆C₃ × ₆C₃ = 20 × 20 = 400

If 4 girls are selected, the number of combinations is:

₆C₂ × ₆C₄ = 15 × 15 = 225

The probability that there are more boys than girls is:

(1 + 36 + 225) / (1 + 36 + 225 + 400 + 225)

= 262 / 887

≈ 0.2954

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A recent article in Business Week listed the "Best Small Companies." We are interested in the current results of the companies'

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Answer:

(i) The estimated regression equation is;

$\hat y$ ≈ 1.6896 + 0.0604·X

The coefficient of 'X' indicates that $\hat y$ increase by a multiple of 0.0604 for each million dollar increase in sales, X

(ii) The estimated earnings for the company is approximately $4.7096 million

(iii) The standard error of estimate is approximately 29.34

The high standard error of estimate indicates that individual mean do not accurately represent the population mean

(iv) The coefficient of determination is approximately 0.57925

The coefficient of determination indicates that the probability of the coordinate of a new point of data to be located on the line is 0.57925

Step-by-step explanation:

The given data is presented as follows;

$\begin{array}{ccc}Sales \ (\$million)&&Earning \ (\$million) \\89.2&&4.9\\18.6&&4.4\\18.2&&1.3\\71.7&&8\\58.6&&6.6\\46.8&&4.1\\17.5&&2.6\\11.9&&1.7\end{array}$

(i) From the data, we have;

The regression equation can be presented as follows;

$\hat y$ = b₀ + b₁·x

Where;

b₁ = The slope given as follows;

$b_1 = \dfrac{\Sigma(x_i - \overline x) \cdot (y_i - \overline y)}{\Sigma(x_i - \overline x)^2}$

b₀ = $\overline y$ - b₁· $\overline x$

From the data, we have;

${\Sigma(x_i - \overline x) \cdot (y_i - \overline y)}$ = 364.05

$\Sigma(x_i - \overline x)^2}$ = 6,027.259

$\overline y$ = 4.2

$\overline x$ = 41.5625

∴ b₁ = 364.05/6,027.259 ≈ 0.06040059005

b₀ = 4.2 - 0.06040059005 × 41.5625 ≈ 1.68960047605 ≈ 1.69

Therefore, we have the regression equation as follows;

$\hat y$ ≈ 1.6896 + 0.0604·X

The coefficient of 'X' indicates that the earnings increase by a multiple of 0.0604 for each million dollar increase in sales

(ii) For the small company, we have;

X = $50.0 million, therefore, we get;

$\hat y$ = 1.6896 + 0.0604 × 50 = 4.7096

The estimated earnings for the company, $\hat y$ = 4.7096 million

(iii) The standard error of estimate, σ, is given by the following formula;

$\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{n - 1}}$

Where;

n = The sample size

Therefore, we have;

$\sigma =\sqrt{\dfrac{6,027.259 }{8 - 1}} \approx 29.34$

The standard error of estimate, σ ≈ 29.34

The high standard error of estimate indicates that it is very unlikely that a given mean value within the data is a representation of the true population mean

(iv) The coefficient of determination (R Square) is given as follows;

$R^2 = \dfrac{SSR}{SST}$