Question

Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. The distribution of pH levels for all community swimming pools in a large county is approximately normal with mean 7.5 and standard deviation 0.2. According to swimming pool studies, the safest pH levels for water in swimming pools are between 7.2 and 7.8. (a) One community swimming pool in the county will be selected at random. What is the probability that the selected pool has a pH level that is not considered safe

Answer 1

Answer:

The  value is  $P(A) = 0.133617$

Step-by-step explanation:

From the question we are told that

  The  mean is  $\mu = 7.5$

  The standard deviation is  $\sigma = 0.2$

  The safest water level is  between  7.2 and  7.8

Generally the probability that the selected pool has a pH level that is not considered safe is mathematically represented as

       $P(A) = 1 - P(7.2 \le X \le 7.8 )$

Here  

      $P(7.2 < X < 7.8 ) = P(\frac{ 7.2 - \mu }{\sigma } < \frac{X - \mu }{ \sigma }$

Generally $\frac{X - \mu }{ \sigma } = Z (The \ standardized \ value \ of X )$

So

 $P(7.2 < X < 7.8 ) = P(\frac{ 7.2 - 7.5 }{0.2 } < Z$    

 $P(7.2 < X < 7.8 ) = P(-1.5 < Z$  

=>  $P(7.2 < X < 7.8 ) = P(Z < 1.5) - P( Z < - 1.5)$

From the z-table the probability of  (Z <  -1.5) and   (  Z <1.5)  are

    $P(Z < 1.5) = 0.93319$

and  

    $P(Z < -1.5) = 0.066807$

So

      $P(7.2 < X < 7.8 ) =0.93319 - 0.066807$

       $P(7.2 < X < 7.8 ) =0.0866383$

So

   $P(A) = 1 - 0.0866383$

=>  $P(A) = 0.133617$