Answer to Question #202735 in Statistics and Probability for nada

Question

Previous record has indicated that the breaking strength of cables used in textile industries is normally distributed with a known variance of 25KN/m2 . A random sample of ten cables is tested resulting in the following yields: 250 225 190 188 210 210 198 230 233 231 Construct a 98% confidence interval on the true average breaking strength.

Answer 1

$s= \sqrt{25}=5\\ n=10\\ \mu = \frac{250 +225 +190 +188 +210 +210 +198 +230 +233 +231}{10}=216.5$

Z-value for 98% CI = 2.33

Confidence interval:

$CI = \mu ± Z \times \frac{s}{\sqrt{n}} \\ CI = 216.5± 2.33 \times \frac{5}{\sqrt{10}} \\ = 216.5± 3.68 \\ 212.82 ≤ \mu ≤ 220.18$