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astraxan [27]
3 years ago
5

In a math class, 12 out of 15 girls are freshmen and 11 out of 15 boys are freshmen. What is the probability that in a randomly

selected group of five students from the class, there will be two freshmen girls and three freshmen boys? Express your answer as a common fraction.
Mathematics
1 answer:
SashulF [63]3 years ago
4 0

Answer:

P(2FG|3FB) = \frac{605}{7917}

Step-by-step explanation:

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter.To calculate combinations the below  formula is used:

nCr = \frac{n!}{r!(n-r)!}

where n represents the total number of items and r represent the number of items being chosen at a time. The '!' represents factorial function which is the product of all integers equal to and less than the given integer. This can be calculated manually using the formula or by using the nCr function on a scientific calculator.

Thus, the number of ways to select 2 freshmen girls (2FG) and 3 freshmen boys (3FB) can be determined. The number of ways to select 5 students (5S) can be determined in the same way.

2FG = 12C2 = \frac{12!}{2!(12-2)!} = \frac{12!}{2!(10)!} = 66

3FB = 11C3 = \frac{11!}{3!(11-3)!} = \frac{11!}{3!(8)!} = 165

5S = 30C5 = \frac{30!}{5!(30-5)!} = \frac{30!}{5!(25)!} = 142506

The probability of selecting 2 freshmen girls (2FG) and 3 freshmen boys (3FB) when selecting 5 students out of 30 is given as below:

P(2FG|3FB) = \frac{2FG*3FB}{5S} = \frac{66*165}{142506} = \frac{605}{7917}

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Answer:

(a)\ \bar x_1 - \bar x_2 = 2.0

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Step-by-step explanation:

Given

n_1 = 60     n_2 = 35      

\bar x_1 = 13.6    \bar x_2 = 11.6    

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Solving (a): Point estimate of difference of mean

This is calculated as: \bar x_1 - \bar x_2

\bar x_1 - \bar x_2 = 13.6 - 11.6

\bar x_1 - \bar x_2 = 2.0

Solving (b): 90% confidence interval

We have:

c = 90\%

c = 0.90

Confidence level is: 1 - \alpha

1 - \alpha = c

1 - \alpha = 0.90

\alpha = 0.10

Calculate z_{\alpha/2}

z_{\alpha/2} = z_{0.10/2}

z_{\alpha/2} = z_{0.05}

The z score is:

z_{\alpha/2} = z_{0.05} =1.645

The endpoints of the confidence level is:

(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}

2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}

2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}

2.0 \± 1.645 * \sqrt{0.0735+0.2571}

2.0 \± 1.645 * \sqrt{0.3306}

2.0 \± 0.9458

Split

(2.0 - 0.9458) \to (2.0 + 0.9458)

(1.0542) \to (2.9458)

Hence, the 90% confidence interval is:

CI =(1.0542,2.9458)

Solving (c): 95% confidence interval

We have:

c = 95\%

c = 0.95

Confidence level is: 1 - \alpha

1 - \alpha = c

1 - \alpha = 0.95

\alpha = 0.05

Calculate z_{\alpha/2}

z_{\alpha/2} = z_{0.05/2}

z_{\alpha/2} = z_{0.025}

The z score is:

z_{\alpha/2} = z_{0.025} =1.96

The endpoints of the confidence level is:

(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}

2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}

2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}

2.0 \± 1.96 * \sqrt{0.0735+0.2571}

2.0 \± 1.96* \sqrt{0.3306}

2.0 \± 1.1270

Split

(2.0 - 1.1270) \to (2.0 + 1.1270)

(0.8730) \to (2.1270)

Hence, the 95% confidence interval is:

CI = (0.8730,2.1270)

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