Question

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.

Answer 1

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  $n = 49$

       The mean is  $\mu = 17ppb$

       The standard deviation is $\sigma = 14 ppb$

Generally the standard error of this measurement is mathematically represented as

       $\sigma_z = \frac{\sigma}{\sqrt{n} }$      

substituting values

      $\sigma_{\= x} = \frac{14}{\sqrt{49} }$  

     $\sigma_{\= x} = 2$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

    $z = \frac{\mu -\sigma }{\sigma_{\= x}}$

     $z = \frac{15-17}{2}$

      $z = -1$

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

      $P(X$