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

Qaulity control plan A calls for rejecting a battery as defective if its weight falls more than 2 standard deviations below the specified mean.
Qaulity control plan B calls for rejecting a battery as defective if its weight falls more than 1.5 interquartile ranges below the lower quartile of the specified population.

a) What proportions of batteries will be rejected by plan A?
I got .025.

b) What is the probability that at least 1 of 2 randomly chosen batteries will be rejected by plan A?
I am completely drawing blank on this one.

c) What proportions of batteries will be rejected by plan B?
I don't know how to do this one either.

9-1

c. Q3-Q1=IQR

-.67-.67=1.34
Z=Q1-1.5IQR
Z=.67-1.5(1.34)
Z=-2.68
P(Z>-2.68)=.0037

To answer these questions, we need to use properties of the normal distribution and the interquartile range.

a) To find the proportion of batteries rejected by plan A, we need to calculate the probability that a randomly chosen battery falls more than 2 standard deviations below the mean.

Since we know that the weights of batteries are normally distributed, we can use the Z-score formula to standardize the values and find the corresponding probabilities in a standard normal distribution table.

The Z-score formula is: Z = (X - μ) / σ

Where:
X = the battery weight
μ = the mean of the battery weights
σ = the standard deviation of the battery weights

Since plan A rejects batteries falling more than 2 standard deviations below the mean, we need to find the probability of Z < -2.

Using a standard normal distribution table or statistical software, we find that the probability P(Z < -2) is approximately 0.025.

So, the proportion of batteries rejected by plan A is approximately 0.025.

b) To find the probability that at least 1 of 2 randomly chosen batteries will be rejected by plan A, we can use the complement rule.

The complement rule states that the probability of an event occurring is equal to 1 minus the probability of the event not occurring.

The probability that at least 1 of the 2 batteries is rejected can be calculated as 1 - P(both batteries not rejected).

For each battery, the probability of not being rejected is 1 - 0.025 = 0.975. Therefore, the probability of both batteries not being rejected is 0.975 * 0.975 = 0.950625.

Using the complement rule, the probability that at least 1 of the 2 batteries is rejected by plan A is approximately 1 - 0.950625 = 0.049375, or about 4.94%.

c) To find the proportion of batteries rejected by plan B, we need to calculate the probability that a randomly chosen battery falls more than 1.5 interquartile ranges below the lower quartile (Q1) of the specified population.

The interquartile range (IQR) represents the spread of the middle 50% of the data and is given by IQR = Q3 - Q1, where Q3 is the upper quartile.

First, we need to find the values of Q1 and Q3. Once we have those, we can calculate 1.5 times the IQR below Q1 and find the probability of a battery falling below that threshold.

Once we know the values of Q1 and Q3, we calculate the IQR = Q3 - Q1 and then calculate the threshold as Q1 - (1.5 * IQR).

After finding the threshold value, we find the probability that a battery falls below that threshold by calculating the area to the left of the threshold in the normal distribution.

Unfortunately, you haven't provided the specific values for mean, standard deviation, Q1, and Q3, so it is not possible to calculate the proportion of batteries that will be rejected by plan B without this information.

a) .0225

b) =1-P(x=0)=.0445
c) I do not know.