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We are looking to find P(X>60 students)
X is normally distributed with mean 50 and standard deviation 5
We need to find the z-score of 60 students
To find the probability of P(Z>2), we can do 1 - P(Z<2)
So we read the probability when Z<2 which is 0.9772, then subtract from one we get 0.0228
The number of students that has score more than 60 is 0.0228 x 1000 = 228 students
X is normally distributed with mean 50 and standard deviation 5
We need to find the z-score of 60 students
To find the probability of P(Z>2), we can do 1 - P(Z<2)
So we read the probability when Z<2 which is 0.9772, then subtract from one we get 0.0228
The number of students that has score more than 60 is 0.0228 x 1000 = 228 students
Answer:
-12
Step-by-step explanation:
Let b = -1 and a = -3
The average rate of change =
f(b) = f(-1) = 3(-1)^2 - 5 = 3 - 5 = -2
f(a) = f(-3) = 3(-3)^2 - 5 =27 - 5 = 22
f(b) - f(a) = -2 - 22 = -24
b - a = -1 + 3 = 2
= -24/2 = -12