Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 7.5 years, and standard deviation of 1.1 years.If 17 items are picked at random, 10% of the time their mean life will be less than how many years?

Answer 1

`a=7.5 -1.28(1.1)/(√(17))= 7.158`

And 7.158 years is the value that accumulates 10% of the data below and 90% of the data above.

Explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

2) Solution to the problem

Let X the random variable that represent the lifespan of a population, and for this case we know the distribution for X is given by:

`X \sim N(7.5,1.1)`

Where
`\mu=7.5` and
`\sigma1.1`

From the central limit theorem we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

10% of the time their mean life will be less than how many years?

So w want to find the value for the average
`\bar X` from the following equation:

`P(\bar X <a)=0.1`

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

`z=(\bar X- \mu)/((\sigma)/(√(n)))`

If we apply this formula we got:

`P(Z< (a- 7.5)/((1.1)/(√(17))))=0.1`

And the value of Z that accumulates 0.1 of the area on the left of the normal standard distribution is
`z=-1.28` so we can find a from the following equation:

`Z=-1.28=(a- 7.5)/((1.1)/(√(17)))`

And we have this:

`a=7.5 -1.28(1.1)/(√(17))= 7.158`

And 7.158 years is the value that accumulates 10% of the data below and 90% of the data above.