Answer:
First, determine how many standard deviations above the mean one would have to be to be in the 75th percentile. This can be found by using a z table and finding the z associated with 0.75. The value of z is 0.674. Thus, one must be .674 standard deviations above the mean to be in the 75th percentile
Answer:
20%
Step-by-step explanation:
61/100, because if it is 0.(value) than it is value / 100 assuming value is 2 digits long. We cannot simplify it further because 61 is prime number.
Let
A = event that the student is on the honor roll
B = event that the student has a part-time job
C = event that the student is on the honor roll and has a part-time job
We are given
P(A) = 0.40
P(B) = 0.60
P(C) = 0.22
note: P(C) = P(A and B)
We want to find out P(A|B) which is "the probability of getting event A given that we know event B is true". This is a conditional probability
P(A|B) = [P(A and B)]/P(B)
P(A|B) = P(C)/P(B)
P(A|B) = 0.22/0.6
P(A|B) = 0.3667 which is approximate
Convert this to a percentage to get roughly 36.67% and this rounds to 37%
Final Answer: 37%
