Question

Question 7.

The lifetimes of a certain brand of battery are assumed to be normally distributed with a mean of 405 hours.
Find the standard deviation of these battery lifetimes, if 90% of these batteries do not achieve a lifetime of 411
hours.
a) 0.2
b) 3.1
c) 3.7
d) 4.7

Answer 1

Answer: D

Explanation:

To solve this problem, we need to use the standard normal distribution and the z-score formula.

First, we need to find the z-score that corresponds to the 90th percentile. We can use a standard normal distribution table or a calculator to find that the z-score is approximately 1.28.

Next, we can use the formula for the z-score:

z = (x - μ) / σ

where x is the value of interest, μ is the mean, and σ is the standard deviation.

We know that x = 411, μ = 405, and z = 1.28. We can rearrange the formula to solve for σ:

σ = (x - μ) / z

Plugging in the values, we get:

σ = (411 - 405) / 1.28

σ = 4.69

Rounding to one decimal place, we get:

σ ≈ 4.7

Therefore, the standard deviation of the battery lifetimes is approximately 4.7 hours, which is option (d).