Question

Suppose that the length of life in hours (x) of a light bulb manufactured by company A is n(800,14400), and the length of life in hours (y) of a light bulb manufactured by company B is N(850,2500). One bulb is randomly selected from each company and is burned until "death."

Answer 1

The mean length of life for light bulbs from company A is
`\( \bar{x} = 800 \)` hours with a standard deviation of
`\( \sigma_x = 120 \)` hours. For company B, the mean length of life is
`\( \bar{y} = 850 \)` hours with a standard deviation of
`\( \sigma_y = 50 \)` hours.

Step-by-step explanation:

The given information indicates that the length of life for light bulbs from company A follows a normal distribution
`\( N(800,14400) \)`, where 800 is the mean
`(\( \bar{x} \))` and 14400 is the variance
`(\( \sigma_x^2 \))`. The standard deviation
`(\( \sigma_x \))` is the square root of the variance, so
`\( \sigma_x = √(14400) = 120 \)` hours.

Similarly, for company B, the length of life follows a normal distribution
`\( N(850,2500) \)`, where 850 is the mean
`(\( \bar{y} \))` and 2500 is the variance
`(\( \sigma_y^2 \))`. The standard deviation
`(\( \sigma_y \))` is the square root of the variance, so
`\( \sigma_y = √(2500) = 50 \)` hours.

In summary, the mean length of life and standard deviation for each company's light bulbs are determined by the given normal distributions. The provided values specify the central tendency and variability of the lengths of life for bulbs from companies A and B.