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A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a standard deviation of 86 minutes with a mean life of 505 minutes. If the claim is true, in a sample of 120 batteries, what is the probability that the mean battery life would differ from the population mean by greater than 16.6 minutes? Round your answer to four decimal places.

User Snor
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2 Answers

6 votes

Final answer:

Using the Central Limit Theorem, the probability that the mean battery life would differ from the population mean by greater than 16.6 minutes is 0.0171.

Step-by-step explanation:

To solve this problem, we can use the Central Limit Theorem. The distribution for the mean length of time 120 batteries last follows a normal distribution with a mean of 505 minutes and a standard deviation of 86 minutes. To find the probability that the mean battery life would differ from the population mean by greater than 16.6 minutes, we need to calculate the z-score and then find the area under the normal curve beyond that z-score.

Step 1: Calculate the standard error (SE) of the distribution of sample means. SE = standard deviation / square root of sample size. SE = 86 / sqrt(120) = 7.857.

Step 2: Calculate the z-score. z = (sample mean - population mean) / SE. z = (16.6 - 0) / 7.857 = 2.11.

Step 3: Find the area under the normal curve beyond the z-score. Using a standard normal distribution table or calculator, we can find that the area to the right of z = 2.11 is approximately 0.0171.

Therefore, the probability that the mean battery life would differ from the population mean by greater than 16.6 minutes is 0.0171.

User NiMux
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4.7k points
2 votes

Answer:

p=0.01740

Step-by-step explanation:

-Let
\bar X denote the mean battery life and that it follows a normal distribution of mean 505 minutes and standard deviation of 86 minutes.

-To find the probability in a sample n=120, we calculate as follows:


z=(\bar X-\mu)/(\sigma/√(n))\\\\=(16.6)/(86/√(120))\\\\=2.1145

The probability is thus:


P(Z>2.1145)=1-0.9826\\\\=0.01740

Hence, the probability is 0.01740

User John Doedoe
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5.1k points