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aleksklad [387]
3 years ago
10

Confidence Interval Application Many people feel that the drying of pavement marking paint is much too slow. You spend several d

ays looking for the fastest drying paint you can find; you plan to measure the time (in seconds) for this paint to dry. From information provided by the paint supplier, you believe the time to dry is normally distributed with a standard deviation of 4 seconds. How many paint samples would you need to test to be able to obtain an estimate of paint drying time that is within 1.5 seconds of the actual mean drying time with a probability of 97%
Mathematics
1 answer:
djyliett [7]3 years ago
7 0

Answer:

The sample size is  n =33

Step-by-step explanation:

From the question we are told that

  The margin of error is E  =  1.5 seconds

   The standard deviation is  s =  4 seconds

 Given that the confidence level is 97% then the level of significance is mathematically represented as

      \alpha =( 100 -97)\%

=>   \alpha  =  0.03

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  2.17

Generally the sample size is mathematically represented as

    n  =[  \frac{Z_{\frac{\sigma }{2 } } *  \sigma }{E} ]^2

=> n  =[  \frac{2.17  * 4 }{1.5} ]^2

=>  n =33

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Answer:

100% probability that the sample mean weight of these 100 bags is less than 18.6 ounces

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}}.

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In this question, we have that:

\mu = 18.5, \sigma = 0.2, n = 100, s = \frac{0.2}{\sqrt{100}} = 0.02

What is the probability that the sample mean weight of these 100 bags is less than 18.6 ounces

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

By the Central Limit Theorem

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

Z = \frac{18.6 - 18.5}{0.02}

Z = 5 has a pvalue of 1

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