Question

A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a standard deviation of 11. The professor has informed us that 7.93 percent of her students received grades of A. What is the minimum score needed to receive a grade of A?

Answer 1

Answer:

88.51 is the minimum score needed to receive a grade of A.          

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 73

Standard Deviation, σ = 11

We are given that the distribution of exam grades is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.0793.

$P( X \geq x) = P( z \geq \displaystyle\frac{x - 73}{11})=0.0793$  

$= 1 -P( z \leq \displaystyle\frac{x - 73}{11})=0.0793$  

$=P( z \leq \displaystyle\frac{x - 73}{11})=0.9207$  

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

$P(z \leq 1.410) = 0.9207$

$\displaystyle\frac{x - 73}{11} = 1.410\\x =88.51$  

Hence, 88.51 is the minimum score needed to receive a grade of A.