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:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the value of x such that the probability is 0.0793.
[tex]P( X \geq x) = P( z \geq \displaystyle\frac{x - 73}{11})=0.0793[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 73}{11})=0.0793 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 73}{11})=0.9207 [/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z \leq 1.410) = 0.9207[/tex]
[tex]\displaystyle\frac{x - 73}{11} = 1.410\\x =88.51[/tex]
Hence, 88.51 is the minimum score needed to receive a grade of A.
