Answer:
The correct solution is "0.0226".
Step-by-step explanation:
The given question seems to be incomplete. Please find below the attachment of the complete query.
According to the question,
Mean
= 29
Standard deviation (s),
= 8
For sample size pf 92,
The standard error will be:
[tex]SE=\frac{s}{\sqrt{N} }[/tex]
[tex]=\frac{8}{\sqrt{92} }[/tex]
[tex]=0.834[/tex]
now,
⇒ [tex]1-P(\frac{-1.9}{0.834} < z < \frac{1.9}{0.834} )[/tex] = [tex]1-P(-2.28<z<2.28)[/tex]
or,
= [tex]1-(2\times P(z<2.28)-1)[/tex]
= [tex]2-2\times P(z<2.28)[/tex]
With the help of table, the normal distribution will be:
= [tex]2-2\times 0.9887[/tex]
= [tex]0.0226[/tex]
