Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 6.7 years, and standard deviation of 1.7 years. The 7% of items with the shortest lifespan will last less than how many years?

Answer 1

4.2 years

Explanation:

If a continuous random variable X is normally distributed with mean μ and variance σ², it is written as:

`\boxed{X \sim\text{N}(\mu,\sigma^2)}`

Given:

• Mean μ = 6.7
• Standard deviation σ = 1.7

Therefore, if the lifespans of an item are normally distributed:

`\boxed{X \sim\text{N}(6.7,1.7^2)}`

where X is the lifespan of the item.

Converting to the Z distribution

`\boxed{\textsf{If }\: X \sim\textsf{N}(\mu,\sigma^2)\:\textsf{ then }\: (X-\mu)/(\sigma)=Z, \quad \textsf{where }\: Z \sim \textsf{N}(0,1)}`

To find the number of years that 7% of the items with the shortest lifespan will last less than, we need to find the value of a for which P(X < a) = 7%:

`\implies \text{P}(X < a) =0.07`

Transform X to Z:

`\text{P}(X < a) = \text{P}\left(Z < (a-6.7)/(1.7)\right)=0.07`

According to the z-tables, when p = 0.07, z = -1.47579106...

`\implies (a-6.7)/(1.7)= -1.47579106...`

`\implies a-6.7= -2.50884480...`

`\implies a=4.19115519...`

`\implies a=4.2\; \sf years`

Therefore, the 7% of items with the shortest lifespan will last less than 4.2 years.