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tankabanditka [31]
3 years ago
11

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of

38443844with a mean life of 997997minutes. If the claim is true, in a sample of 7373batteries, what is the probability that the mean battery life would be greater than 981.7981.7minutes
Mathematics
1 answer:
emmainna [20.7K]3 years ago
3 0

Answer:

z = \frac{981.7 -997}{\frac{62}{\sqrt{73}}}= -2.108

And we can use the normal standard distribution table or excel and with the complement ruler we got:

P(z>-2.108) =1- P(z

Step-by-step explanation:

We know the following info given:

\mu = 997 represent the true mean

\sigma = \sqrt{3844}= 62 represent the population deviation

n = 73 represent the sample size selected

Since the sample size is large enough (n>30) we can use the central limit theorem and the sample mean would have the following distribution:

\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})

And for this case we want to find this probability:

P(\bar X >981.7)

And we can use the z score formula given by:

z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{981.7 -997}{\frac{62}{\sqrt{73}}}= -2.108

And we can use the normal standard distribution table or excel and with the complement ruler we got:

P(z>-2.108) =1- P(z

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