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 .

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:

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

Note: we considered that the margin is ±0.03.

Then, the probability is:

the equation for this situation is y=122+45. to solve the equation you would add 45 and 122 to get a total of 167lbs. in conclusion, grace's weight eight weeks ago was 167lbs.