Answer: 0.0918, it is not unusual.
Step-by-step explanation:
Given : The length of time a person takes to decide which shoes to purchase is normally distributed with a mean of 8.54 minutes and a standard deviation of 1.91.
i.e. [tex]\mu=8.54[/tex] minutes and [tex]\sigma= 1.91[/tex] minutes
Let x denotes the length of time a person takes to decide which shoes to purchase.
Formula : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
Then, the probability that a randomly selected individual will take less than 6 minutes to select a shoe purchase will be :-
[tex]\text{P-value=}P(x<6)=P(\dfrac{x-\mu}{\sigma}<\dfrac{6-8.54}{1.91})\\\\\approx P(z<1.33)=1-P(z<1.33)\ \ \ [\becaus\ P(Z<-z)=1-P(Z<z)]\\\\=1-0.9082\ \ [\text{By using z-value}]=0.0918[/tex]
Thus , the required probability = 0.0918
Since, P-value (0.0918) >0.05 , it means this outcome is not unusual.
[Note : When a outcome is unusual then the probability of its happening is less than or equal to 0.05. ]
