Answer:
0.0082
Step-by-step explanation:
Data provided in the question:
Mean = 10
Variance = 25
Standard deviation = [tex]\sqrt{\text{variance}}
[tex]=\sqrt{25}=5 [/tex]
n=36
Given [tex]P(12 \leq \bar{x})[/tex]
[tex]=P(\bar{x} \geq 12)[/tex]
[tex]=P\left(\frac{\bar{x}-\mu_{x}}{\frac{\sigma}{\sqrt{n}}} \geq \frac{12-10}{\left(\frac{5}{\sqrt{36}}\right)}\right)[/tex]
[tex]=P(z \geqslant 2 \cdot 4)[/tex]
[tex]=1-P(z<2 \cdot 4)[/tex]
By using z-table we get,
[tex]=1-0.9918[/tex]
= 0.0082.