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A scientist studying water quality measures the lead level in parts per billion (ppb) at each of 49 randomly chosen locations along a water line. Suppose that the lead levels across all the locations on this line are strongly skewed to the right with a mean of 17 ppb and a standard deviation of 14 ppb. Assume that the measurements in the sample are independent. What is the probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb? Choose 1 answer: A) Plæ <15) = 0.02 B) Plū<15) – 0.16 C) Plē <15) 0.30 D) Plö < 15) – 0.44 E) We cannot calculate this probability because the sampling distribution is not normal.

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okpalawalter8

Complete Question

The complete qustion is shown on the first uploaded image

Answer:

The correct option is B

Step-by-step explanation:

From the question we are told that

      The sample size  is  [tex]n = 49[/tex]

       The mean is  [tex]\mu = 17ppb[/tex]

       The standard deviation is [tex]\sigma = 14 ppb[/tex]

Generally the standard error of this measurement is mathematically represented as

       [tex]\sigma_z = \frac{\sigma}{\sqrt{n} }[/tex]      

substituting values

      [tex]\sigma_{\= x} = \frac{14}{\sqrt{49} }[/tex]  

     [tex]\sigma_{\= x} = 2[/tex]ppb

Now the probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb represented as P(X < 15 )

Next is to find the z value

    [tex]z = \frac{\mu -\sigma }{\sigma_{\= x}}[/tex]

     [tex]z = \frac{15-17}{2}[/tex]

      [tex]z = -1[/tex]

Now checking the z-table for the z-score of  -1 we have  

      [tex]P(X<15) = P(Z < -1 )= 0.16[/tex]

                       

       

     

Ver imagen okpalawalter8

Using the normal distribution and the central limit theorem, it is found that there is a 0.16 = 16% probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb, hence option B is correct.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex], considering a sample size of at least 30 for a skewed variable.

In this problem:

  • Mean of 17 ppb, hence [tex]\mu = 17[/tex].
  • Standard deviation of 14 ppb, hence [tex]\sigma = 14[/tex].
  • Sample of 49, hence [tex]n = 49, s = \frac{14}{\sqrt{49}} = 2[/tex]

The probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb is the p-value of Z when X = 15, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{15 - 17}{2}[/tex]

[tex]Z = -1[/tex]

[tex]Z = -1[/tex] has a p-value of 0.16

0.16 = 16% probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb, hence option B is correct.

For more on the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213

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