In a study of the health effects of cigarettes, a random sample of 32 filtered cigarettes was obtained and the tar content was

Question

In a study of the health effects of cigarettes, a random sample of 32 filtered cigarettes
was obtained and the tar content was measured. The sample has a mean of 19.2 mg.
Given that the tar content of cigarettes have a mean of 20.1 mg and a standard
deviation of 3.15 mg, what is the probability that 32 filtered cigarettes have a mean of
19.2 mg or less?
Round answer to 4 decimal places.

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Thiên Thanh 4 years 2021-08-22T10:30:16+00:00 1 Answers 37 views 0

Answers ( )

  1. minhkhue
    0
    2021-08-22T10:31:30+00:00
    Reply

    Answer:

    0.0526 = 5.26% probability that 32 filtered cigarettes have a mean of 19.2 mg or less.

    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 normal distributions can be 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.

    Given that the tar content of cigarettes have a mean of 20.1 mg and a standard deviation of 3.15 mg

    This means that \mu = 20.1, \sigma = 3.15

    Sample of 32 filtered cigarettes

    This means that n = 32, s = \frac{3.15}{\sqrt{32}} = 0.5568

    What is the probability that 32 filtered cigarettes have a mean of 19.2 mg or less?

    This is the pvalue of Z when X = 19.2. So

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

    By the Central Limit Theorem

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

    Z = \frac{19.2 - 20.1}{0.5568}

    Z = -1.62

    Z = -1.62 has a pvalue of 0.0526

    0.0526 = 5.26% probability that 32 filtered cigarettes have a mean of 19.2 mg or less.

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