Respuesta :
Answer:
Value of 26.93 will be exceeded 85% of the time.
Step-by-step explanation:
We are given that a random variable is normally distributed with a mean of 25 and a standard deviation of 5.
Let X = a random variable
So, X ~ N([tex]\mu=25,\sigma^{2} =5^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean
[tex]\sigma[/tex] = standard deviation
Now, we have to find that value which will be exceeded 85% of the time, i.e.;
P(X > [tex]x[/tex]) = 0.85
1 - P(X [tex]\leq x[/tex]) = 0.85
P(X [tex]\leq x[/tex]) = 0.15
P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq \frac{x-25}{5}[/tex] ) = 0.15
P(Z [tex]\leq \frac{x-25}{5}[/tex] ) = 0.15
Now, in the z table the critical value of [tex]x[/tex] whose less than area is 0.15 is given as 0.3853, i.e;
[tex]\frac{x-25}{5}[/tex] = 0.3853
[tex]x-25=0.3853 \times 5[/tex]
[tex]x[/tex] = 25 + 1.9265 = 26.93
Therefore, value of 26.93 will be exceeded 85% of the time.
