A manufacturer knows that their items have a normally distributed lifespan, with a mean of 5.6 years, and standard deviation of 1.6 years.
If 25 items are picked at random, 1% of the time their mean life will be less than how many years?
Give your answer to one decimal place.
from a z-score table
1% of the population is approx. 2.33 s.d. below the mean
2.33 * 1.6 = ?
To find the answer to this question, we need to use the concept of the sampling distribution of the mean. According to the central limit theorem, for a large enough sample size, the sampling distribution of the mean will be approximately normal, regardless of the shape of the population distribution.
In this case, we are given that the population follows a normal distribution with a mean (μ) of 5.6 years and a standard deviation (σ) of 1.6 years. We are asked to find the value below which the mean lifespan of 25 randomly selected items will fall 1% of the time.
First, we calculate the standard deviation of the sampling distribution of the mean, also known as the standard error (SE), using the formula:
SE = σ / √n
where σ is the population standard deviation, and n is the sample size.
In this case, σ = 1.6 years and n = 25, so:
SE = 1.6 / √25
SE = 1.6 / 5
SE = 0.32 years
Next, we need to find the z-score corresponding to the desired percentile. Since we want to find the value below which the mean lifespan will fall 1% of the time, we need to find the z-score for the 1st percentile, which is equivalent to the z-score for the 0.01 level of significance (α = 0.01).
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to the 0.01 level of significance is approximately -2.33.
Finally, we can calculate the value below which the mean lifespan of 25 randomly selected items will fall 1% of the time by using the formula:
X = μ + (z * SE)
where X is the value we are looking for, μ is the population mean, z is the z-score, and SE is the standard error.
Substituting the given values, we get:
X = 5.6 + (-2.33 * 0.32)
X = 5.6 - 0.7456
X ≈ 4.8544
Therefore, the mean lifespan of 25 randomly selected items will be less than approximately 4.9 years 1% of the time.
you can play around with Z-table stuff at
http://davidmlane.com/hyperstat/z_table.html