Answer:
The appropriate answer is option b (0.0228).
Step-by-step explanation:
The given values are:
Random sample,
[tex]n=100[/tex]
Average length of time,
[tex]\bar{x}=3.1 \ minutes[/tex]
Waiting time,
[tex]\mu=3 \ minutes[/tex]
Standard deviation,
[tex]\sigma=0.5[/tex]
Now,
The test statistics will be:
⇒ [tex]z=\frac{(\bar{x}-\mu)}{\frac{\sigma}{\sqrt{n} } }[/tex]
On substituting the values, we get
⇒ [tex]=\frac{(3.1-3)}{\frac{0.5}{\sqrt{100} } }[/tex]
⇒ [tex]=\frac{0.1}{0.05}[/tex]
⇒ [tex]=2[/tex]
hence,
⇒ [tex]P(z >2 ) = 1 - P(z <2 )[/tex]
By using the table,
⇒ [tex]=1-0.9772[/tex]
⇒ [tex]=0.0228[/tex]
