Answer to Question #151789 in Statistics and Probability for Quero

Question

A factory manufactures light bulbs for distribution under 2 different brands. Brand A bulbs have an average life of 1800 hours with a standard deviation of 240 hours; Those of mark B have an average life of 1450 hours with a standard deviation of 150 hours. A sample of 250 light bulbs is taken from each brand. Describe completely this situation, define all the data, and explain the method and all steps of computation. How likely is the average lifespan of Brand A bulbs to be at least 400 hours longer than that of Brand B bulbs?

Answer 1

Let's decribe each sample:

$\overline{X}$ - sample mean

A:

$\mu_{\overline{X}_A}$ = 1800

$\sigma_{\overline{X}_A}$ = $\sigma/\sqrt{n}$ = $240/\sqrt{250}=15.1789$

B:

$\mu_{\overline{X}_B}$ = 1450

$\sigma_{\overline{X}_B}$ = $\sigma/\sqrt{n}$ = $150/\sqrt{250}=9.48683$

Let's decribe the difference between two means:

$\mu_{{\overline{X}_A}-{\overline{X}_B}} = 1800-1450=350$

$\sigma_{{\overline{X}_A}-{\overline{X}_B}}=17.8997,\sqrt{\sigma_{\overline{X}_A}^2+\sigma_{\overline{X}_B}^2}=\sqrt{15.1789^2+9.48683^2}=17.8997$

Finally, we can compute z-score:

z=$\frac{400-\mu_{({\overline{X}_A}-{\overline{X}_B})}}{\sigma_{({\overline{X}_A}-{\overline{X}_B})}}=\frac{400-350}{17.8997}=2.79334$

P(x>z) = 0.0026083

Answer: 0.0026083