Q:

A randomly selected customer support wait time​ (from a normal​ distribution) is calculated to be 1.8 standard deviations above its mean. What is the probability that another randomly selected customer wait time from the distribution will be less than 1.8 standard deviations from the​ mean? Round to four decimal places as needed.

Accepted Solution

A:
Answer: 0.9640Step-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]