Answer:
The scores above 1.08 standard deviation from the mean are publicly recognized.
Step-by-step explanation:
We are given the following information in the question:
Mean = μ
Standard Deviation = σ
We are given that the distribution of score 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 0.14
[tex]P( X > x) = P( z > \displaystyle\frac{x - \mu}{\sigma})=0.14[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - \mu}{\sigma})=0.14[/tex]
[tex]=P( z \leq \displaystyle\frac{x - \mu}{\sigma})=0.86[/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - \mu}{\sigma} = 1.08\\\\x =\mu + 1.08\sigma[/tex]
Thus, scores above 1.08 standard deviation from the mean are publicly recognized.