Question 4.30: Let X denote the number of flaws in a 1 in. length of copper......
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.
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?
| x | P(X = x) |
| 0 | 0.48 |
| 1 | 0.39 |
| 2 | 0.12 |
| 3 | 0.01 |
The population mean number of flaws is 𝜇 = 0.66, and the population variance is \sigma^{2} = 0.5244. (These quantities can be calculated using Equations 3.13 and 3.14 in Section 3.3.)
\mu _{X} = \sum\limits_{x}{xP(X = x)} (3.13)
\sigma ^{2}_{X} = \sum\limits_{x}{(x\ −\ \mu _{X})^{2}P(X = x)} (3.14)
Let X_{1},…, X_{100} denote the number of flaws in the 100 wires sampled from this population. We need to find P(\overline{X} < 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 \overline{X} ∼ N(0.66,\ 0.005244). The z-score of 0.5 is therefore
\overline{X} ∼ N (μ, \frac{σ^2}{n}) approximately (4.33)
z = \frac{0.5\ −\ 0.66}{\sqrt{0.005244}} = −2.21
From the z table, the area to the left of −2.21 is 0.0136. Therefore P(\overline{X} < 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.
