Answer:
The lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 500
Standard Deviation, σ = 50
We are given that the distribution of test score is a bell shaped distribution that is a normal distribution.
Formula:
![z_{score} = \displaystyle\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z_%7Bscore%7D%20%3D%20%5Cdisplaystyle%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
We have to find the value of x such that the probability is 0.06.
P(X > x) = 6% = 0.06
Calculation the value from standard normal z table, we have,
![P(z < 1.555) = 0.94](https://tex.z-dn.net/?f=P%28z%20%3C%201.555%29%20%3D%200.94)
Hence, the lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.
Answer:
27/462
OR
.058
Step-by-step explanation:
the probability that the first student, selected at random, is a nursing student is 3/22.
the probability that the second student, selected at random from the remaining students, is an accounting major is 9/21.
the probability the first student chosen is a nursing student and the second student chosen is an accounting major is therefore 3/22 * 9/21 = 27/462.
Answer:
the answers is 3 over 2 in fraction form
It is less than. Any negative number compared to a positive number will be less than.