Respuesta :
Answer:800
Step-by-step explanation:
[tex]Mean(\mu )=\$ 4300 [/tex]
and [tex]P(X>6140)=0.0107[/tex]
[tex]P(X>x_0)=0.0107[/tex]
[tex]P\left ( \frac{X-\mu }{\sigma}<\frac{x-\mu }{\sigma} \right )=0.9893[/tex]
[tex]P\left ( z<\frac{x-\mu }{\sigma} \right )=0.9893[/tex]
[tex]\frac{x-\mu }{\sigma}=InvNormal(0.9893)[/tex]
Using Z table to get InvNormal(0.9893)=2.30
[tex]\frac{6140-4300}{\sigma}=2.30[/tex]
[tex]\sigma =\frac{6140-4300}{2.30}=800[/tex]