Answer:
c) (40+60+25)/200 or 63%
Step-by-step explanation:
n= 200 students
Did Well on the Midterm and Studied for the Midterm = 75
Did Well on the Midterm and Went Partying = 40
Did Poorly on the Midterm and Studied for the Midterm = 25
Did Poorly on the Midterm and Went Partying = 60
The number of students that did poorly on the midterm or went partying the weekend before the midterm is given by the sum of all students who did poorly to all students who went partying minus the number of students who did Poorly on the Midterm and Went Partying:
The probability that a randomly selected student did poorly on the midterm or went partying the weekend before the midterm is given by:
-4x + 6y = 12
x + 2y = -10
First solve for x in the second equation
x + 2y = -10
x = -10 - 2y
Now we have a value for x so we can substitute it into the other equation
-4 (-10 - 2y) + 6y = 12
Now solve for y
40 + 8y + 6y = 12
40 + 14y = 12
14y = -28
y = -2
Now we have a value for y that we can plug into one of the original equations so we can solve for x
x + 2y = -10
x + 2(-2) = -10
x - 4 = -10
x = -6
Your solution set is
(-6, -2)
X=-11 :) let me know if you need more help!
Answer:
a. 2.28%
b. 30.85%
c. 628.16
d. 474.67
Step-by-step explanation:
For a given value x, the related z-score is computed as z = (x-500)/100.
a. The z-score related to 700 is (700-500)/100 = 2, and P(Z > 2) = 0.0228 (2.28%)
b. The z-score related to 550 is (550-500)/100 = 0.5, and P(Z > 0.5) = 0.3085 (30.85%)
c. We are looking for a value b such that P(Z > b) = 0.1, i.e., b is the 90th quantile of the standard normal distribution, so, b = 1.281552. Therefore, P((X-500)/100 > 1.281552) = 0.1, equivalently P(X > 500 + 100(1.281552)) = 0.1 and the minimun SAT score needed to be in the highest 10% of the population is 628.1552
d. We are looking for a value c such that P(Z > c) = 0.6, i.e., c is the 40th quantile of the standard normal distribution, so, c = -0.2533471. Therefore, P((X-500)/100 > -0.2533471) = 0.6, equivalently P(X > 500 + 100(-0.2533471)), and the minimun SAT score needed to be accepted is 474.6653
