Question

The pH measurements of water specimens from various locations along a given river basin are Normally distributed with mean 8 and standard deviation 0.3.
a). What is, approximately, the probability that the pH measurement of a randomly selected water specimen is greater than 8.2?
b). What is, approximately, the probability that the pH measurement of a randomly selected water specimen is a value between 7.5 and 8.2?
c) Find the first quartile of the distribution of the pH measurements. i.e., Find the pH value such that three-quarters of the pH measurements in this river basin are greater than this value.

Answer 1

Answer:

a

  $P(X > 8.2) =0.25239$

b

  $P(7.2 < X < 8.2 ) = 0.6999$

c

    $x = 7.7978$

Step-by-step explanation:

From the question we are told that

   The  mean is  $\mu = 8$

   The standard deviation is  $\sigma = 0.3$

probability that the pH measurement of a randomly selected water specimen is greater than 8.2 is mathematically evaluated as

    $P(X > 8.2) = P(\frac{X - \mu }{\sigma } > \frac{ 8.2 - 8 }{0.3} )$

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

   $P(X > 8.2) = P(Z >0.667 )$

From the z table  the area under the normal curve to the left corresponding to  0.667  is

       $P(Z >0.667 ) = 0.25239$

So

     $P(X > 8.2) =0.25239$

Generally the probability that the  that the pH measurement of a randomly selected water specimen is a value between 7.5 and 8.2 is mathematically represented as

        $P(7.5 < X < 8.2 ) = P( \frac{7.5 - 8}{0.3} < \frac{X - \mu }{\sigma} < \frac{8.2- 8}{0.3} )$

=>    $P(7.5 < X < 8.2 ) = P(-1.667 < Z < 0.667 )$

=>    $P(7.5 < X < 8.2 ) = P(Z< 0.667 ) - P( Z < -1.667 )$

From the z table the area under the normal curve to the left corresponding to  -2.667 and  0.667   is

       $P( Z < -1.667 ) =0.047757$

and  

        $P(Z< 0.667 ) = 0.74761$

So  

    $P(7.2 < X < 8.2 ) = 0.74761 - 0.047757$

=> $P(7.2 < X < 8.2 ) = 0.6999$

Generally 25% of the pH measurements is below the first quartile of the distribution

  Hence the probability of the obtaining a value below the  first quartile is mathematically represented as

     $P(X < x ) = P( \frac{X - \mu }{\sigma } < \frac{x-8}{0.3} ) = 0.25$

=>   $P( Z < \frac{x-8}{0.3} ) = 0.25$

Generally from the normal distribution table the critical value corresponding to 0.25  is  

      $z = -0.674$

So

       $\frac{x-8}{0.3} = -0.674$

=>   $x = 7.7978$