Respuesta :
Answer:
Grade B score:
[tex]76 \leq x \leq 89[/tex]
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 73.3
Standard Deviation, σ = 9.7
We are given that the distribution of score on test is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
B: Scores below the top 5% and above the bottom 62%
We have to find the value of x such that the probability is 0.62
[tex]P( X > x) = P( z > \displaystyle\frac{x - 73.3}{9.7})=0.62[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 73.3}{9.7})=0.62 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 73.3}{9.7})=0.38 [/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 73.3}{9.7} = 0.305\\\\x = 76.26[/tex]
We have to find the value of x such that the probability is 0.05
[tex]P(X < 0.95) = \\\\P( X < x) = P( z < \displaystyle\frac{x - 73.3}{9.7})=0.95[/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 73.3}{9.7} = 1.645\\\\x = 89.26[/tex]
Thus, the numerical value of score to achieve grade B is
[tex]76 \leq x \leq 89[/tex]