State importance of central limit theorem From random sample of 40 divers what is the probability that mean time they spend in water is between 60minutes and 90 minutes if the standard deviation is 2 minutes

Population mean: $\mu = 75$ minutes (midpoint of 60 and 90 minutes).
Population standard deviation: $\sigma = 2$ minutes.
Sample size: $n=40$.

Formula: $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$.
Substitute: $\sigma_{\bar{X}} = \frac{2}{\sqrt{40}}$.
Simplify: $\sqrt{40} = 2\sqrt{10} \approx 6.3246$, so $\sigma_{\bar{X}} \approx \frac{2}{6.3246} \approx 0.3162$ minutes.
Exact: $\sigma_{\bar{X}} = \frac{1}{\sqrt{10}}$ minutes.

Lower bound formula: $z_1 = \frac{60 - 75}{\sigma_{\bar{X}}}$.
Substitute: $z_1 = \frac{-15}{0.3162} \approx -47.43$.

Upper bound formula: $z_2 = \frac{90 - 75}{\sigma_{\bar{X}}}$.
Substitute: $z_2 = \frac{15}{0.3162} \approx 47.43$.

where $Z \sim N(0,1)$.
For $|z| > 5$, $\Phi(z) \approx 1$, so $P \approx 1$ (or 100%).

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