Answer: 0.9640
Step-by-step explanation:
Let [tex]\mu[/tex] be the mean wait time and [tex]\sigma[/tex] be the standardr deviation.
Let x be the random variable that represents the wait time .
Given : A randomly selected customer support wait time (from a normal distribution) is calculated to be 1.8 standard deviations above its mean.
then , [tex]x=\mu+1.8\sigma[/tex]
For z -score ,
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
[tex]z=\dfrac{\mu+1.8\sigma-\mu}{\sigma}\\\\\Rightarrow\ z=\dfrac{1.8\sigma}{\sigma}=1.8[/tex]
By using standard normal distribution table , the probability that another randomly selected customer wait time from the distribution will be less than 1.8 standard deviations from the mean :-
[tex]P(z<1.8)=0.9640696\approx0.9640[/tex]
