Question

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?

Answer 1

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 2

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 %