Suppose that the grades of Business Mathematics and Statistic module is modeled well by a normal probability distribution with mean (212) and standard deviation (122). Let
X be the random variable representing this distribution. Find two symmetric values "a"
and "b" such that Probability [ a

a = -2.57 and b = 2.57

Step-by-step explanation:

Explanation:

Given mean of the Population = 212

Standard deviation of the Population = 122

Let  X be the random variable of the Normal distribution

Given P( a ≤ z ≤b) = 0.99

  put a=-b

         P( -b ≤ z ≤b) = 0.99

   ⇒     |A( b) - A( -b)| =0.99

    ⇒     | A( b) + A( b)| =0.99  

   ⇒       2 |A (b)| = 0.99

     ⇒     | A(b)| = 0.495

From normal table find value in areas

          b = 2.57 ( see in  normal table)

         Given a =-b

                   a = - 2.57

location: point 0

step-by-step explanation:

well, -8 to 10 on a number line is 18 spaces.

1/9 of 18 is 2.

so, 4/9 of 18 must be 8.

8 spaces from -8 (a) is 0.

so, 4/9 of the way from -8 to 10 is 0!