Respuesta :
Answer:
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 )%
