A population of plastic chairs in a factory has a weight’s mean of 1.5 kg and a standard deviation of 0.1 kg . Suppose a sample of size 100

Question

A population of plastic chairs in a factory has a weight’s mean of 1.5 kg and a standard deviation of 0.1 kg . Suppose a sample of size 100 is selected and X is used to estimate u . What is the probability that the sample mean will be within +0.02 of the population mean ?​

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Euphemia 4 years 2021-07-25T17:50:54+00:00 1 Answers 8 views 0

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  1. vankhanh
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    2021-07-25T17:52:30+00:00
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    Answer:

    0.9544 = 95.44% probability that the sample mean will be within +0.02 of the population mean.

    Step-by-step explanation:

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

    Normal probability distribution

    When the distribution is normal, we use 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 question, we have that:

    \mu = 1.5, \sigma = 0.1, n = 100, s = \frac{0.1}{\sqrt{100}} = 0.01

    What is the probability that the sample mean will be within +0.02 of the population mean?

    Sample mean between 1.5 – 0.02 = 1.48 kg and 1.5 + 0.02 = 1.52 kg, which is the pvalue of Z when X = 1.52 subtracted by the pvalue of Z when X = 1.48. So

    X = 1.52

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

    By the Central Limit Theorem

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

    Z = \frac{1.52 - 1.5}{0.01}

    Z = 2

    Z = 2 has a pvalue of 0.9772

    X = 1.48 ​

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

    Z = \frac{1.48 - 1.5}{0.01}

    Z = -2

    Z = -2 has a pvalue of 0.0228

    0.9772 – 0.0228 = 0.9544

    0.9544 = 95.44% probability that the sample mean will be within +0.02 of the population mean.

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