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

Importance of the Central Limit Theorem

The Central Limit Theorem (CLT) states that the distribution of the sample mean $\bar{X}$ from any population with mean $\mu$ and standard deviation $\sigma$ is approximately normal $N\left(\mu, \frac{\sigma^2}{n}\right)$ for large sample size $n$ (typically $n \geq 30$), regardless of the population's distribution shape.

This is crucial because it enables the use of normal probabilities and z-scores for inference on $\mu$ from sample data, even for skewed or unknown population distributions.

Step 1: Identify parameters.
Population mean: $\mu = 75$ minutes (midpoint of 60 and 90 minutes).
Population standard deviation: $\sigma = 2$ minutes.
Sample size: $n=40$.

Step 2: Compute standard error of the mean (by CLT).
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.

Step 3: Standardize the bounds to z-scores (by CLT, $\bar{X} \approx N(75, 0.3162^2)$).
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: $z_2 = \frac{15}{0.3162} \approx 47.43$.

Step 4: Compute probability.
$P(60 < \bar{X} < 90) = P(-47.43 < Z < 47.43)$
where $Z \sim N(0,1)$.
For $|z| > 5$, $\Phi(z) \approx 1$, so $P \approx 1$ (or 100%).

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