Beginning statistics, simple confidence interval problem

Question

I am in a first statistics class and we have been covering confidence intervals. This seems like a confidence interval problem but I'm a bit thrown by it. The problem states:

"A machine fills $100$-lb bags of dry concrete mix. Because of slight difficult-to-correct imprecision in the machine dispensing mechanism, the actual weight of the mix in a $100$-lb bag varies from bag to bag, and hence can be considered a random variable. It has been determined that the weight of a bag is approximately normally distributed with a standard deviation of $ \sigma = 0.34 $ lb.

The mean of the weight of a bag, however, can be set by the operator.

a) Suppose each bag is considered “within specification” if its weight is within $100 \pm 0.75$ lb. An operator sets the mean weight at $100.20$ lb. What percentage of the bags filled by the machine is considered “within specification”?"

I'm not sure how to interpret what's going on. What are they asking me? If the mean weight can be set by the operator then clearly we aren't interested in an estimate of that. I'm very lost as to how to approach this.

score 1 · Accepted Answer · answered Dec 05 '13 at 01:27

1

The question asks what is the probability that $99.25 \le X\le 100.75$ if $X$ is normal, mean $100.2$, standard deviation $0.34$.

This is $\Pr(X\le 100.75) -\Pr(X\le 99.25)$. Each of the required probabilities is straightforward to compute using software, or a table of the standard normal.

Remark: I would not call this a confidence interval problem, it is a probability calculation.

answered Dec 05 '13 at 01:27

André Nicolas

507,029

I see it now. I was trying to use a hammer on something that's not a nail. Thank you. – R R Dec 05 '13 at 01:31
You are welcome. Good way to put it. – André Nicolas Dec 05 '13 at 01:32

Beginning statistics, simple confidence interval problem

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