The ap statistics exam, like all other ap exams, is graded on a 5, 4, 3, 2, 1, 0 basis. many colleges require a 4 or a 5 to grant credit. in 1999, 25,240 students took the exam, and 30.4% of these received either a 4 or a 5. assume that the raw scores were approximately normally distributed with a mean of 60 and a standard deviation of 19. what is the minimum score a student must receive to earn a 4 or a 5?

a.70
b.71
c.72
d.69
e.68

A minimum score of 70 is required for a student to earn a 4 or a 5.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 60

Standard Deviation, σ = 19

We are given that the distribution of raw scores is a bell shaped distribution that is a normal distribution.

Formula:

Top 30.4% students received either a 4 or a 5. Let x denotes the score that separates top 30.4% students from lowest 69.6% students

We have to find the value of x such that the probability is 0.304

P(X > x)  

Calculation the value from standard normal z table, we have,  

A minimum score of 70 is required for a student to earn a 4 or a 5.

0.85 or 85%

step-by-step explanation:

1) divide 23 by 27

2) round the decimal to the nearest 10

3) so your answer would be 0.85 or 85%

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18