I believe it is 805.2. It is not 671
In addition to what Paul in CT said, 13 is a member of the natural numbers, whole numbers, rational numbers, real numbers, complex numbers, and algebraic numbers.
A is a subset of B is all the elements of A are elements of B
The standard deviation of the frequency distribution is 5.54
<h3>How to determine standard deviation?</h3>
The table of values is given as:
x f(x)
0-3 13
4-7 13
8-11 10
12-15 11
16-19 0
20-23 3
Rewrite the table by calculating the class midpoints:
x f(x)
1.5 13
5.5 13
9.5 10
13.5 11
17.5 0
21.5 3
Start by calculating the mean using:

This gives

Evaluate

The standard deviation is then calculated as:

So, we have:

Evaluate

Solve

Hence, the standard deviation is 5.54
Read more about standard deviation at:
brainly.com/question/15858152
#SPJ1
All you have to so is divide 1968 by 4.
1968 ÷ 4 = 496.5 = 1/4
Now multiply 496.5 by 3 = 1, 489.5 = 3/4
I hope this helps! :_
Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:

The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:
X = 590:


Z = 0.76
Z = 0.76 has a p-value of 0.7764.
X = 400:


Z = -0.89
Z = -0.89 has a p-value of 0.1867.
0.7764 - 0.1867 = 0.5897 = 58.97%.
58.97% of students would be expected to score between 400 and 590.
More can be learned about the normal distribution at brainly.com/question/27643290
#SPJ1