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 )%        

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.

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