A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a standard deviation of 6

Question

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a standard deviation of 6262 minutes with a mean life of 606606 minutes. If the claim is true, in a sample of 9999 batteries, what is the probability that the mean battery life would be greater than 619619 minutes? Round your answer to four decimal places.

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Isabella 1 month 2021-10-22T00:38:22+00:00 1 Answer 0 views 0

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    2021-10-22T00:39:34+00:00

    Answer:

    0.0183 = 1.83% probability that the mean battery life would be greater than 619 minutes

    Step-by-step explanation:

    To solve this question, we need to understand the normal probability distribution and the central limit theorem.

    Normal probability distribution

    Problems of normally distributed samples are solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

    Central Limit Theorem

    The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

    For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

    In this problem, we have that:

    \mu = 606, \sigma = 62, n = 99, s = \frac{62}{\sqrt{99}} = 6.23

    What is the probability that the mean battery life would be greater than 619 minutes?

    This is 1 subtracted by the pvalue of Z when X = 619. So

    Z = \frac{X - \mu}{\sigma}

    By the Central Limit Theorem

    Z = \frac{X - \mu}{s}

    Z = \frac{619 - 606}{6.23}

    Z = 2.09

    Z = 2.09 has a pvalue of 0.9817

    1 – 0.9817 = 0.0183

    0.0183 = 1.83% probability that the mean battery life would be greater than 619 minutes

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