Question

your turn
boxes of cruncho cereal have a mean mass of 323 grams with a standard deviation of 20 grams.

1. for a random sample of 50 boxes, what interval captures 99.7% of all means under the normal distribution?

Explanation:

Step1: Calculate the standard error

The formula for standard error \(SE=\frac{\sigma}{\sqrt{n}}\), where \(\sigma = 20\) grams and \(n = 50\).
\(SE=\frac{20}{\sqrt{50}}\approx\frac{20}{7.07}\approx2.83\)

Step2: Determine the z - score

For a \(99.7\%\) confidence interval in a normal distribution, the z - score \(z = 3\) (based on the empirical rule: about \(99.7\%\) of the data lies within \(z=\pm3\) in a standard normal distribution).

Step3: Calculate the margin of error

The margin of error \(ME = z\times SE\). Substituting \(z = 3\) and \(SE\approx2.83\), we get \(ME=3\times2.83 = 8.49\)

Step4: Find the interval

The lower limit \(= \mu - ME\) and the upper limit \(=\mu+ME\), where \(\mu = 323\) grams.
Lower limit: \(323 - 8.49=314.51\)
Upper limit: \(323 + 8.49 = 331.49\)

Answer:

The interval \((314.51,331.49)\) grams.