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Julli [10]
2 years ago
11

At your​ school, 20​% of the class are marketing majors. You are randomly assigned to two partners in your statistics class. ​a)

What is the probability that the first partner will be a marketing​ major? ​b) What is the probability that the first partner​ won't be a marketing​ major? ​c) What is the probability that both will be marketing​ majors? ​d) What is the probability that at least one will be a marketing​ major?
Mathematics
1 answer:
Digiron [165]2 years ago
5 0

Answer:

a) 20%

b) 80%

c) There is 4% probability that both will be marketing majors.

d) There is a 36% probability that at least one will be a marketing​ major.

Step-by-step explanation:

At your​ school, 20​% of the class are marketing majors.

This means that for each student, there is a 20% that he is a marketing major and an 80% chance that he is not a marketing major.

a) What is the probability that the first partner will be a marketing​ major?

The probabilities for each student are independent, so it is 20%.

​b) What is the probability that the first partner​ won't be a marketing​ major?

80%

c) What is the probability that both will be marketing​ majors?

For each one, it is 20%. So:

P = 0.2*0.2 = 0.04

There is 4% probability that both will be marketing majors.

​d) What is the probability that at least one will be a marketing​ major?

That is one of them or both. For both, we found in c) that it is 4%.

For one of them, there are two cases. Either the first partner is and the second is not, or the other way. So, the probability that at least one will be a marketing major is

P = 0.04 + 2*0.2*0.8 = 0.36

There is a 36% probability that at least one will be a marketing​ major.

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<h2>Answer:</h2>

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<h2>Step-by-step explanation:</h2>

We know that the data points given by (x,y) are in direct variation if:

\dfrac{y}{x}=k  for each (x,y)

We are given data as:

Caleb’s Earnings Hours(x)        12          15         18         21

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So, we could see the ratio as:

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Hence, we could see that each ratio are not equal.

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Twenty students from Sherman High School were accepted at Wallaby University. Of those students, eight were offered military sch
Shalnov [3]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let w be the subscript for scores of students with military scholarship and o be the subscript for scores of students without military scholarship.

Therefore, the population means would be μw and μo.

The random variable is xw - xo = difference in the sample mean scores of students with military scholarships and students without.

For students with military scholarship,

n = 8

Mean = (850 + 925 + 980 + 1080 + 1200 + 1220 + 1240 + 1300)/8

Mean = 1099.375

Standard deviation = √(summation(x - mean)/n

Summation(x - mean) = (850 - 1099.375)^2 + (925 - 1099.375)^2 + (980 - 1099.375)^2 + (1080 - 1099.375)^2 + (1200 - 1099.375)^2 + (1220 - 1099.375)^2 + (1240 - 1099.375)^2 + (1300 -1099.375)^2 = 191921.875

Standard deviation = √(191921.875/8 = 154.89

For students without military scholarship,

n = 12

Mean = (820 + 850 + 980 + 1010 + 1020 + 1080 + 1100 + 1120 + 1120 + 1200 + 1220 + 1330)/12

Mean = 1073.83

Summation(x - mean) = (820 - 1073.83)^2 + (850 - 1073.83)^2 + (980 - 1073.83)^2 + (1010 - 1073.83)^2 + (1020 - 1073.83)^2 + (1080 - 1073.83)^2 + (1100 - 1073.83)^2 + (1120 - 1073.83)^2 + (1120 - 1073.83)^2 + (1200 - 1073.83)^2 + (1220 - 1073.83)^2 + (1330 - 1073.83)^2 = 238199.4268

Standard deviation = √(238199.4268/12 = 140.89

We would set up the hypothesis.

The null hypothesis is

H0 : μw = μo H0 : μw - μo = 0

The alternative hypothesis is

Ha : μw ≠ μo Ha : μw - μo ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(xw - xo)/√(sw²/nw + so²/no)

From the information given,

xw = 1099.375

xo = 1073.83

sw = 154.89

so = 140.89

nw = 8

no = 12

t = (1099.375 - 1073.83)/√(154.89²/8 + 140.89²/12)

t = 0.37

The formula for determining the degree of freedom is

df = [sw²/nw + so²/no]²/(1/nw - 1)(sw²/nw)² + (1/no - 1)(so²/no)²

df = [154.89²/8 + 140.89²/12]²/(1/8 - 1)(154.89²/8)² + (1/12 - 1)(140.89²/12)² = 21650688.37/1533492.15

df = 14

We would determine the probability value from the t test calculator. It becomes

p value = 0.72

Since the level of significance of 0.05 < the p value of 0.72, we would not reject the null hypothesis.

Therefore, these data do not provide convincing evidence of a difference in SAT scores between students with and without a military scholarship.

Part B

The formula for determining the confidence interval for the difference of two population means is expressed as

z = (xw - xo) ± z ×√(sw²/nw + so²/no)

For a 95% confidence interval, the z score is 1.96

xw - xo = 1099.375 - 1073.83 = 25.55

z√(sw²/nw + so²/no) = 1.96 × √(154.89²/8 + 140.89²/12) = 1.96 × √2998.86 + 1654.17)

= 133.7

The confidence interval is

25.55 ± 133.7

6 0
2 years ago
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