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:
<u><em>Step(i):-</em></u>
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](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7Bx-mean%7D%7B%5Cfrac%7BS.D%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20%3D%20%5Cfrac%7B15-17%7D%7B%5Cfrac%7B14%7D%7B%5Csqrt%7B49%7D%20%7D%20%7D%20%3D%20-1)
<u><em>Step(ii):</em></u>-
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
<u><em>Final answer:-</em></u>
The probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb
P(x⁻< 15) = 0.1587