Answer:
Grade D score:
[tex]69 \leq x \leq 74[/tex]
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 78.8
Standard Deviation, σ = 7.1
We are given that the distribution of scores on test is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
D: Scores below the top 77% and above the bottom 9%
We have to find the value of x such that the probability is 0.77
[tex]P( X > x) = P( z > \displaystyle\frac{x - 78.8}{7.1})=0.77[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 78.8}{7.1})=0.77[/tex]
[tex]=P( z \leq \displaystyle\frac{x - 78.8}{7.1})=0.23[/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 78.8}{7.1} =-0.739\\\\x = 73.55[/tex]
We have to find the value of x such that the probability is 0.09
[tex]P(X < 0.09) = \\\\P( X < x) = P( z < \displaystyle\frac{x - 78.8}{7.1})=0.09[/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 78.8}{7.1} = -1.341\\\\x = 69.27[/tex]
Thus, the numerical value of score to achieve grade D is
[tex]69 \leq x \leq 74[/tex]