Central Limit Theorem Math Example 1

Follow the full solution, then compare it with the other examples linked below.

Example 1

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A highly skewed population (times between bus arrivals) has ฮผ=15\mu=15 min and ฯƒ=8\sigma=8 min. For samples of n=64n=64, describe the shape, mean, and SD of the sampling distribution of Xห‰\bar{X}, and find P(Xห‰<14)P(\bar{X} < 14).

Solution

  1. 1
    CLT: despite skewed population, with n=64โ‰ฅ30n=64 \geq 30, Xห‰\bar{X} is approximately normally distributed
  2. 2
    Mean: ฮผXห‰=15\mu_{\bar{X}} = 15 min; SE: ฯƒXห‰=864=88=1\sigma_{\bar{X}} = \frac{8}{\sqrt{64}} = \frac{8}{8} = 1 min
  3. 3
    Xห‰โˆผN(15,1)\bar{X} \sim N(15, 1)
  4. 4
    P(Xห‰<14)=P(Z<14โˆ’151)=P(Z<โˆ’1)=0.1587P(\bar{X} < 14) = P(Z < \frac{14-15}{1}) = P(Z < -1) = 0.1587

Answer

Xห‰โˆผN(15,1)\bar{X} \sim N(15, 1); P(Xห‰<14)โ‰ˆ0.159P(\bar{X} < 14) \approx 0.159 despite non-normal population.
The CLT's power: even with a skewed population, sample means become normally distributed with large enough n. This allows us to use normal-distribution methods (z-scores, standard tables) for any population shape, which is why CLT is central to statistical inference.

About Central Limit Theorem

For sufficiently large sample size (nโ‰ฅ30n \geq 30 as a rule of thumb), the sampling distribution of the sample mean is approximately normal with mean ฮผ\mu and standard deviation ฯƒn\frac{\sigma}{\sqrt{n}}, regardless of the shape of the population distribution.

Learn more about Central Limit Theorem โ†’

More Central Limit Theorem Examples

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