Question

Suppose that the raw daily oxygen purities delivered by an air-products supplier have a standard deviation LaTeX: \sigma\approx.1 σ ≈ .1 (percent), and it is plausible to think of daily purites as independent random variables. Approximate the probability that the sample mean LaTeX: \frac{ }{X} X of n = 25 delivered purities falls within .03 (percent) of the raw daily purity mean, LaTeX: \mu μ .

Answer 1

Answer:

There is a probability of 86.6% that the sample mean falls within 0.03 percent of the raw purity mean.

Step-by-step explanation:

We have a population standard deviation of σ ≈ 0.1.

We have a sample of size n=25.

Then, we have a sampling distribution, which has a standard deviation for the sample mean that is:

$\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{0.1}{\sqrt{25}}=\dfrac{0.1}{5}=0.02$

Now, we can calculate a z-score for a deviation of 0.03 percent from the mean as:

$z=\dfrac{X-\mu}{\sigma}=\dfrac{0.03}{0.02}=\dfrac{0.03}{0.02}=1.5$

Note: we considered that the margin is ±0.03.

Then, the probability is:

$P(|X-\mu|<0.03\%)=P(|z|<1.5)=0.866$