Let X denote the number of flaws in a 1 in. length of copper wire. The probability mass function of X is presented in the following table.

Question

[latex]\begin{array}{cc}\hline x & P ( X=x ) \\hline 0 & 0.48 \1 & 0.39 \2 & 0.12 \3 & 0.01 \\hline\end{array}[/latex]

One hundred wires are sampled from this population. What is the probability that the average number of flaws per wire in this sample is less than 0.5?

Accepted Answer

The population mean number of flaws is μ = 0.66, and the population variance is σ² = 0.5244. (These quantities can be calculated using Equations 3.13 and 3.14 in Section 3.3.) Let [latex]X_1, .... , X_{1oo}[/latex] denote the number of flaws in the 100 wires sampled from this population. We need to find P([latex]\bar{X}[/latex] < 0.5). Now the sample size is n = 100, which is a large sample. It follows from the Central Limit Theorem (expression 4.33) that [latex]\bar{X}[/latex] ∼ N (0.66, 0.005244). The z-score of 0.5 is therefore

[latex]\bar{X}\sim N\left(\mu, \frac{\sigma^2}{n} ight) \quad ext{approximately}[/latex] (4.33)

[latex]\mu_X=\sum_x x P(X=x)[/latex] (3.13)

[latex]\sigma_X^2=\sum_x\left(x-\mu_X ight)^2 P(X=x)[/latex] (3 .1 4)

[latex]z=\frac{0.5-0.66}{\sqrt{0.005244}}=-2.21[/latex]

From the z table, the area to the left of - 2.21 is 0.0136.Therefor P([latex]\bar{X}[/latex] < 0.5) = 0.0136, so only 1.36% of samples of size 100 will have fewer than 0.5 flaws per wire. See Figure 4.17.

Note that in Example 4.30 we needed to know only the mean and variance of the population, not the probability mass function.

Question 4.30: Let X denote the number of flaws in a 1 in. length of copper...

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