Question

The tensile strength for a type of wire is normally distributed with unknown mean μ and unknown variance σ₂. Six pieces of wire were randomly selected from a large roll; Yi​, the tensile strength for portion i, is measured for i=1,2,⋯,6. The population mean μ and variance σ₂ can be estimated by yˉ​ and s₂, respectively. Because σyˉ​2​=σ₂/n, it follows that σyˉ​2​ can be estimated by s₂/n. Find the appropriate probability that yˉ​ will be within 2s/n of the true population mean μ.

Answer 1

The probability that
`\( \bar{y} \)` will be within
`\( (2s)/(√(n)) \)` of the true population mean
`\( \mu \)` is approximately 0.9545.

Step-by-step explanation:

The Central Limit Theorem states that for a sufficiently large sample size, the distribution of the sample mean
`\( \bar{y} \)` approaches a normal distribution. The standard error of the sample mean
`\( \sigma_{\bar{y}} \)` is estimated by
`\( (s)/(√(n)) \)`.

To find the probability that
`\( \bar{y} \)` is within
`\( (2s)/(√(n)) \) of \( \mu \)`, we look for the area under the normal curve within two standard errors from the mean. This is approximately 0.9545, which can be obtained from standard normal distribution tables or statistical software.

Therefore, there is a 95.45% probability that
`\( \bar{y} \)` falls within
`\( (2s)/(√(n)) \) of \( \mu \)`.