No one in this world can solve this ap stat question..

Test Preparation AP Test Preparation
1

<p>Here we go… </p>

<p>A company is considering implementing one of two quality control plans for monitoring the weights of automiile batteries that it manufactures. If the manufacturing process is working properly, the battery weights are approximately normally distributed with a specified mean and standard deciation.</p>

<p>Quality control plan A calls for rejecting a battery as defective if its weight falls more that 2 standard deiations below the specified mean.</p>

<p>Quality control plan B calls for rejecting a battery as defective if its weight falls more that 1.5 interquartile ranges below tha lower quartile of the specified population.</p>

<p>Assume the manufacturing process is under control.</p>

<p>a. What proportion of batteries will be rejected by plan A?</p>

<p>b. What proportion of batteries will be rejected by plan B?</p>

<p>I think no one in this world can solve this problem…I am struggling for
two hours and couldnt solve it…AP spat is so hard…</p>

<p>Can anyone solve it??</p>

2

<p>The key is that the populations are normally distrubuted…so use the fact that in a normal distribution, 68% is within one standard deviation of the mean, and 95% is within two.</p>

3

<p>a.) According to the standard deviation model, only ~2.275% of all in the population will be less than 2 standard deviations below the specified mean. 2.275/100 is 91/4000
b.) You would need more information- the lower quartile, the upper quartile, and the values of those below the lower quartile</p>

4

<p>Q3-Q1=IQR
Q3=-.67
Q1=.67
IQR=1.34
z=Q1-1.5IQR
z=-.67-1.5(1.34)
z=-2.68
P(z<-2.68)=.0037</p>

<p>.0037 of the batteries will be rejected by Plan B.</p>

5

<p>jc’s part B is correct</p>

<p>for part A, you know by the Empirical Rule that 95% of all products lie between 2 standard deviations from the mean, and thus 5% symmetrically lie outside, meaning 2.5% lie below 2 standard devations and 2.5% lie above</p>

<p>so…the answer is 2.5% (or .025)</p>