# Thread: Help! - Probability Problem

1. ## Help! - Probability Problem

I need some help with this one:

A company is considering implementing one of two quality control plans for monitoring the weights of automobile batteries it manufactures. If the manufacturing process is working properly, the battery weights are approximately normally distributed with a specific mean and standard deviation.

Quality control plan A calls for rejecting a battery as defective if its weight falls more than 2 standard deviations below the specificed mean.

Quality control plan B calls for rejecting a a battery as defective if its weight falls more than 1.5 interquartile ranges below the lower quartile of the specified population.

Assume the manufacturing process is under normal control.

a. What proportion of batteries will be rejected by plan A?

b. What is the probability that at least 1 of 2 randomly selected batteries will be rejected by plan A?

c. What proportion of batteries will be rejected by plan B?
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a. Since anything below 2 st. dev will be rejected, I look up 2 on the Z table and find the p value, which is .0228?

b. .5 is the sample value, .0228 is the population proportion. So I find a Z-value by doing .5 - .0228/st. dev. But what would I use for the stand. dev since its not given?

c. No idea how to do this... I know IQR measures middle 50% of data.. so 1.5 * .5 = .75.. but where do you go from there?

Im really confused and unsure on what Ive done.. Any help would be much appreciated!

2. A is correct

for B you should simply use the principles of independent events and treat it as an "or" so if the batteries are labeled as A and B the you want

P(A is rejected, OR B is rejected)

for C determine the z-scores of the quartiles adn use those to determine the z-score of the lower quartile, now take the interquartile range (in terms of z-scores) and subtract 1.5 times the IQR from the lower quartile value. you will have a z-score for the rejction value and you can read the probability from there.

jerry

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