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 ppb17 and a standard deviation of 14ppb. Assume that the measurements in the sample are independent.
Required:
What is the probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb?

Answer 1

Answer:

0.16

Step-by-step explanation:

Answer 2

Answer:

The probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb

P(x⁻< 15) = 0.1587

Step-by-step explanation:

Step(i):-

Given that the size of the sample 'n' =49

Mean of the Population = 17ppb

The standard deviation of the population = 14ppb

Let 'X' be the random variable in a normal distribution

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{15-17}{\frac{14}{\sqrt{49} } } = -1$

Step(ii):-

The probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb

P(x⁻< 15) = P(Z<-1) = 1-P(Z>-1)

                             = 1-(0.5+A(-1))

                             = 0.5 - A(1)

                            = 0.5-0.3413

                            = 0.1587

Final answer:-

The probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb

P(x⁻< 15) = 0.1587