Using the normal distribution, it is found that 0.0764 = 7.64% of teenagers who will have waist sizes greater than 31 inches.
<h3>Normal Probability Distribution</h3>
In a normal distribution with mean 
and standard deviation 
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of
.
- The standard deviation is of
.
The proportion of teenagers who will have waist sizes greater than 31 inches is <u>1 subtracted by the p-value of Z when X = 31</u>, hence:
has a p-value of 0.9236.
1 - 0.9236 = 0.0764.
0.0764 = 7.64% of teenagers who will have waist sizes greater than 31 inches.
More can be learned about the normal distribution at brainly.com/question/24663213
Statement that describes the difference that exists between an online identity and a real-life one is Online identities are more flexible.
Online identities can be regarded as the identity that is been presented on the internet and this is usually different from the real life identity.
This is because on internet different means can be used to have fake identity and this is different from real world.
We can conclude that difference between online identity and a real-life one is flexibility.
CHECK THE COMPLETE RELATED QUESTION BELOW;
What BEST describes the difference between an online identity and a real-life one?
Online identities are decided for you.
Online identities are short-lived.
Online identities are more flexible.
Online identities are forced on you
Learn more about online identity at:
brainly.com/question/7331447
Answer:
6.2125 g
Explanation:
6.2125 g of Na₂SO₄ is needed to prepare 350 mL of a solution having a sodium ion concentration of 0.125 M.
