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?

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 :)

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slicergiza slicergiza · Answer 2 · 2019-06-18T07:56:35-04:00

Answer:

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

Step-by-step explanation:

Given,

Mean, \(\mu\) = 600,

Standard deviation, \(\sigma\) = 20,

Using the equation,

\(z-score=\frac{x-\mu}{\sigma}\)

Z-score for the data 580 = \(\frac{580-600}{20}\)

\(=\frac{-20}{20}\)

\(=-1\)

While, the z-score for the data 620 = \(\frac{620-600}{20}\)

\(=\frac{20}{20}\)

\(=1\)

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 %