Answer:
[tex]\displaystyle\frac{x - 78}{8.6} = -0.440[/tex]
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 78
Standard Deviation, σ = 8.6
We are given that the distribution 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 0.33
[tex]P( X > x) = P( z > \displaystyle\frac{x - 78}{8.6})=0.33[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 78}{8.6})=0.33 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 78}{8.6})=0.67[/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 78}{8.6} = -0.440\\\\x = 74.22[/tex]
which is the required equation.
Thus, values greater than equal to 74.22 above which 33% of the score fall.
