f the life, in months, of a certain type of car battery is normally distributed with a mean of 88months and a standard deviation of 9 months, what should be the guarantee period if the companywants less than 5% of the batteries to fail while under warranty?
Answers:
A)Less than 16.13 monthsB)Less than 73.15 monthsC)Less than 64.78 monthsD)Less than 102.85 months
The Correct one is B, but I don't know how to get it. Can someone show me a step by step solution please.
this like your 1st question except you're going the other way with the z-score table
they give you the population percentage and want the z-score
... the table shows -1.645
... multiply by 9 to find the number of months less than 88
same webpage I gave you before:
https://www.jiskha.com/display.cgi?id=1524703735
click on : value from an area
enter:
area = .05
mean = 88
SD = 9
click on "below" to get 73.193
I meant to give you this page:
http://davidmlane.com/normal.html
Thank you both!
To find the guarantee period, we need to determine a value such that less than 5% of the batteries will fail within that period.
Since the life of the car battery is normally distributed, we can use the z-score formula to find the appropriate value. The z-score formula is:
z = (x - μ) / σ
Where:
- x is the value we want to find (guarantee period)
- μ is the mean (88 months)
- σ is the standard deviation (9 months)
We need to find the z-score value associated with a 5% failure rate. This means that we want to find z such that 5% of the area lies to the right of it.
Using a standard normal distribution table or a calculator, we can find the z-score associated with a 5% failure rate, which turns out to be approximately -1.645 (rounded to three decimal places).
To find the guarantee period (x), we can rearrange the z-score formula:
x = μ + z * σ
Substituting the values we know:
- μ = 88 months
- σ = 9 months
- z = -1.645
x = 88 + (-1.645) * 9
x ≈ 73.15
Therefore, the guarantee period should be less than 73.15 months.
Hence, the correct answer is B) Less than 73.15 months.