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PLSSSSSSSSSSS HEEEEEELP????The mean of a set of normally distributed data is 600 with a standard deviation of 20. What percent of the data is between 580 and 620?

Respuesta :

ashlynjayde2003
z = (X-Mean)/SD 
z1 = (580-600)/20 = - 1 
z2 = (620-600)/20 = + 1 
According to the Empirical Rule 68-95-99.7 
Mean +/- 1*SD covers 68% of the values 
Therefore, required percent is 68%
hopes this help :) :D :)

Answer:

The percentage of the data between 580 and 620 is 68.26 %

Step-by-step explanation:

Given,

Mean, [tex]\mu[/tex] = 600,

Standard deviation, [tex]\sigma[/tex] = 20,

Using the equation,

[tex]z-score=\frac{x-\mu}{\sigma}[/tex]

Z-score for the data 580 = [tex]\frac{580-600}{20}[/tex]

[tex]=\frac{-20}{20}[/tex]

[tex]=-1[/tex]

While, the z-score for the data 620 = [tex]\frac{620-600}{20}[/tex]

[tex]=\frac{20}{20}[/tex]

[tex]=1[/tex]

Using normal distribution table,

P(580 < z) = 0.1587

Also, P(620 > z) = 0.8413

Hence, the percent of the data is between 580 and 620 = P(620 > z)  - P(580 < z)

= 0.8413  - 0.1587

= 0.6826

= 68.26 %