Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 7.5 years, and standard deviation of 1.8 years.

If you randomly purchase one item, what is the probability it will last longer than 12 years?

Answer 1

Solution :

Given that,

mean = = 7.5

standard deviation = = 1.8

P(x >12 ) =

= 1 - P[(x -) / < (12-7.5) /1.8 ]

= 1 - P(z <2.5 )

Using z table

= 1 -  0.9938

probability= 0.0062

Answer 2

We can use the Z-score formula to calculate the probability that the item will last longer than 12 years:

Z = (X - μ) / σ

where X is the value we are interested in (12 years in this case), μ is the mean (7.5 years), and σ is the standard deviation (1.8 years).

Z = (12 - 7.5) / 1.8 Z = 2.5 / 1.8 Z = 1.39

Using a standard normal distribution table or calculator, we can find that the probability of a Z-score of 1.39 or greater is approximately 0.0823.

Therefore, the probability that a randomly purchased item will last longer than 12 years is about 0.0823 or 8.23%.