Answer:
The right solution is "68%".
Step-by-step explanation:
The empirical rule is:
[tex]X\sim N(8.43, 1.5)[/tex]
According to the question,
= [tex]P(6.93< \mu < 9.93)[/tex]
= [tex]P(\frac{6.93-8.43}{1.5} < \frac{\mu -8.43}{1.5} < \frac{9.93-8.43}{1.5} )[/tex]
= [tex]P(z(1)-P(z(-1))[/tex]
= [tex]68[/tex] (%)
Thus the above is the right solution.
