Question

a manufacturer knows that their items have a normally distributed lifespan, with a mean of 2.2 years, and standard deviation of 0.7 years. if you randomly purchase 16 items, what is the probability that their mean life will be longer than 2 years?

Answer 1

The probability that the mean lifespan of 16 items is longer than 2 years is approximately 87.08%, given a normally distributed population with a mean of 2.2 years and a standard deviation of 0.7 years.

To find the probability that the mean lifespan of 16 items is longer than 2 years, we can use the Central Limit Theorem. The distribution of sample means for a sufficiently large sample from a population with any shape of distribution approaches a normal distribution.

The mean of the sample mean is the same as the population mean, which is 2.2 years. The standard deviation of the sample mean (also known as the standard error) is given by the population standard deviation divided by the square root of the sample size.

Standard error (SE) =
`\( (0.7)/(√(16)) = (0.7)/(4) = 0.175 \)` years.

Now, we want to find the probability that the sample mean is greater than 2 years. We can convert this to a z-score using the formula:

`\[ Z = ((X - \mu))/(SE) \]`

where X is the value we're interested in,
`\( \mu \)` is the mean, and SE is the standard error.

`\[ Z = ((2 - 2.2))/(0.175) \]`

Calculating this gives the z-score, and then we can find the probability using a standard normal distribution table or calculator.

`\[ Z \approx ((-0.2))/(0.175) \]\[ Z \approx -1.1429 \]`

Using a standard normal distribution table or calculator, the probability that a standard normal random variable is greater than -1.1429 is approximately 0.8708.

So, the probability that the mean lifespan of 16 items is longer than 2 years is approximately 0.8708 or 87.08%.