Answer: 2.14 %
Explanation:
Given : pH measurements of a chemical solutions have
Mean :
![\mu=6.8](https://img.qammunity.org/2020/formulas/mathematics/college/43kvbbf9hz4jbox5a2fsu41fy46avgqdw8.png)
Standard deviation :
![\sigma=0.02](https://img.qammunity.org/2020/formulas/mathematics/college/s7wt8chhs2d3sibl0och1oren0fj7xwo5a.png)
Let X be the pH reading of a randomly selected customer chemical solution.
We assume pH measurements of this solution have a nearly symmetric/bell-curve distribution (i.e. normal distribution).
The z-score for the normal distribution is given by :-
![z=(x-\mu)/(\sigma)](https://img.qammunity.org/2020/formulas/mathematics/high-school/10fia1p0qwvlz4zhb867kzy3u7bscognwz.png)
For x = 6.74
![z=(6.74-6.8)/(0.02)=-3](https://img.qammunity.org/2020/formulas/mathematics/college/sjpnzw1qpdbofgivsoyc5ke8bq5x8dqawx.png)
For x = 6.76
![z=(6.76-6.8)/(0.02)=-2](https://img.qammunity.org/2020/formulas/mathematics/college/48uoabb07gpswg4rnbc4bc9ysubz6iakjh.png)
The p-value =
![P(6.74<x<6,76)=P(-3<z<-2)](https://img.qammunity.org/2020/formulas/mathematics/college/6p93o428c0f8jlymieh3idmdbqecbv17av.png)
![P(z<-2)-P(z<-3)=0.0227501- 0.0013499=0.0214002\approx0.0214](https://img.qammunity.org/2020/formulas/mathematics/college/104nhi5h10m7fneqy9c1ji0c696u8wxa8s.png)
In percent,
![0.0214*=2.14\%](https://img.qammunity.org/2020/formulas/mathematics/college/2nadxwewkx0o0l7b1vkq9csqlxa3abk09x.png)
Hence, the percent of pH measurements reading below 6.74 OR above 6.76 = 2.14%