Suppose the scores on a test given to all juniors in a school district are normally distributed with a mean of 74 and a standard deviation of 8. On a separate sheet of paper, draw a Normal Curve and label the mean, standard deviations, and the associated percentages. Find the percent of juniors who score no more than 90.

PLEASE HELP A

97.5%

Step-by-step explanation:

Solution:-

- Denote a random variable,

           X: Scores of all juniors in a school district centralized test.

- The random variable ( X ) follows normal distribution with the corresponding parameters:

                         X ~ Norm ( μ , σ^2 )

Where,     μ = Mean score

                σ = standard deviation of scores secured

- The given parameters for the normal distribution are:

                        X ~ Norm ( 74 , 8^2 )

- To draw a Normal curve we need to draw a bell shaped curve and annotate the following descriptions:

      Mean ( μ ) : The vertical center-line that bifurcates the normal curve

      1st standard deviation ( μ ± σ ) : First small division to the left and right about the mean ( μ ). [ 74 - 8 , 74 + 8 ] = [ 66 , 82 ]

      2nd standard deviation ( μ ± 2σ ) : Second small division to the left and right about the mean ( μ ). [ 74 - 16 , 74 + 16 ] = [ 58 , 90 ]

      3rd standard deviation ( μ ± 3σ ) : Third small division to the left and right about the mean ( μ ) - tailed. [ 74 - 24 , 74 + 24 ] = [ 50 , 98 ]

- Mark the associated percentage of scores that lies between 1st, 2nd and 3rd standard deviations from the mean ( μ ). Apply the Empirical rule of statistics. Which states:

    p (  μ - σ  ,  μ + σ ) = p ( 66 , 82 ) = 67 %

    p (  μ - 2σ  ,  μ + 2σ ) = p ( 58 , 90 ) = 95 %

    p (  μ - 3σ  ,  μ + 3σ ) = p ( 50 , 98 ) = 99.7 %

- See the attachment for the complete diagram.

- To determine the percentage of students who scored no more than 90 on the test.

- Employ the use of standardizing the required probability by using the following relation:

          p ( X < x ) = p ( Z < [ (x - μ) / σ ] )

          p ( X < 90 ) = p ( Z < [ (90 - 74) / 8 ] )

                             = p ( Z < [ (90 - 74) / 8 ] )

                             = p ( Z < 2 )

- We will employ the use of Empirical rule of second deviation ( μ ± 2σ ) to evaluate the required percentage:

       p (  μ - 2σ  < X <  μ + 2σ ) = p ( 58 , 90 ) = 95 %

       1 - p ( 58 < X < 90 ) = 1 - 0.95 = 0.05

       p ( X > μ + 2σ ) = p ( X > 90 ) = [ 1 - p ( 58 < X < 90 ) ] / 2

                                                      = [ 1 - 0.95 ] / 2

                                                      = 0.05 / 2

                                                      = 0.025

Hence,

     p ( X < 90 ) = p ( Z < 2 ) = 1 - p ( X > 90 )

                                            = 1 - 0.025

     Answer                          = 0.975 ( 97.5 )%        

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step-by-step explanation:

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