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3 years ago

Please help!!!!

Mathematics

1 answer:

Svet_ta [14]3 years ago

3 0

5 would be the best answer choice

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3 years ago

A college job placement office collected data about students’ GPAs and the salaries they earned in their first jobs after gradua

frez [133]

Answer:

X is the GPA

Y is the Salary

Standard deviation of X is 0.4

Standard deviation of Y is 8500

E(X)=2.9

E(Y)=47200

We are given that The correlation between the two variables was r = 0.72

a) $y = a+bx$

$b = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2} = \frac{r \times \sqrt{var(X) \times Var(Y)}}{Var(X)} = \frac{0.72 \times \sqrt{0.4^2 \times 8500^2}}{0.4^2} = 15300$

$a=y-bx = 47200-(15300 \times 29) = 2830$

So, slope = 15300

Intercept = 2830

So, equation : $y = 2830+15300x$

b) Your brother just graduated from that college with a GPA of 3.30. He tells you that based on this model the residual for his pay is -$1880. What salary is he earning?

$y = 2830+15300 \times 3.3 = 53320$

Observed salary = Residual + predicted = -1860+53320 = 51440

c)) What proportion of the variation in salaries is explained by variation in GPA?

The proportion of the variation in salaries is explained by variation in GPA = $r^2 = (0.72)^2 =0.5184$

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2 years ago