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

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

Explanation:

Given,

Mean,
`\mu` = 600,

Standard deviation,
`\sigma` = 20,

Using the equation,

`z-score=(x-\mu)/(\sigma)`

Z-score for the data 580 =
`(580-600)/(20)`

`=(-20)/(20)`

`=-1`

While, the z-score for the data 620 =
`(620-600)/(20)`

`=(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 %