Central Limit Theorem Math Example 2

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Example 2

hard
A fair die (μ=3.5, σ=1.71) is rolled n=100n=100 times. By CLT, find the approximate probability that the sample mean is between 3.3 and 3.7.

Solution

  1. 1
    CLT: Xˉ∼N(3.5,1.71100)=N(3.5,0.171)\bar{X} \sim N\left(3.5, \frac{1.71}{\sqrt{100}}\right) = N(3.5, 0.171)
  2. 2
    P(3.3<Xˉ<3.7)=P(3.3−3.50.171<Z<3.7−3.50.171)=P(−1.17<Z<1.17)P(3.3 < \bar{X} < 3.7) = P\left(\frac{3.3-3.5}{0.171} < Z < \frac{3.7-3.5}{0.171}\right) = P(-1.17 < Z < 1.17)
  3. 3
    P(Z<1.17)≈0.879P(Z < 1.17) \approx 0.879; P(Z<−1.17)≈0.121P(Z < -1.17) \approx 0.121
  4. 4
    P(−1.17<Z<1.17)=0.879−0.121=0.758P(-1.17 < Z < 1.17) = 0.879 - 0.121 = 0.758

Answer

P(3.3<Xˉ<3.7)≈0.758P(3.3 < \bar{X} < 3.7) \approx 0.758. About 75.8% of samples will average between 3.3 and 3.7.
CLT transforms an integer-valued die distribution into a continuous normal distribution for the sample mean. This demonstrates the remarkable universality of the CLT — the average of many dice rolls follows the normal curve regardless of the original distribution.

About Central Limit Theorem

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

Learn more about Central Limit Theorem →

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