Sure, let's break down the problem into a step-by-step solution:
The given standard deviation formula is:
[tex]\[
s=\sqrt{\frac{(18-19.3)^2+(11-19.3)^2+(22-19.3)^2+(26-19.3)^2}{3}}
\][/tex]
### (a) How many data values are there?
The data set consists of 4 values:
- [tex]\( 18 \)[/tex]
- [tex]\( 11 \)[/tex]
- [tex]\( 22 \)[/tex]
- [tex]\( 26 \)[/tex]
Thus, [tex]\( \boxed{4} \)[/tex] data values.
### (b) What are the data values?
The data values are:
- [tex]\( 18 \)[/tex]
- [tex]\( 11 \)[/tex]
- [tex]\( 22 \)[/tex]
- [tex]\( 26 \)[/tex]
Thus, [tex]\( \boxed{18, 11, 22, 26} \)[/tex].
### (c) What is the range of the data?
The range of a data set is the difference between the maximum and the minimum values.
[tex]\[
\text{Range} = \max(\text{data}) - \min(\text{data}) = 26 - 11 = 15
\][/tex]
So, the range of the data is [tex]\( \boxed{15} \)[/tex].
### (d) What is the mean of the data set?
The mean of the data set is given:
[tex]\[
\text{Mean} = 19.3
\][/tex]
Thus, the mean of the data set is [tex]\( 19.3 \)[/tex].
### (e) Compute the standard deviation for this data set.
The calculation for the standard deviation using the given formula involves the following steps:
1. Calculate the deviations from the mean for each data point.
2. Square each deviation.
3. Sum the squared deviations.
4. Divide by [tex]\( n-1 \)[/tex] (where [tex]\( n \)[/tex] is the number of data values, 4 in this case).
5. Take the square root of the result.
The given formula:
[tex]\[
s=\sqrt{\frac{(18-19.3)^2+(11-19.3)^2+(22-19.3)^2+(26-19.3)^2}{3}}
\][/tex]
Substituting the values into the formula, we get:
[tex]\[
s=\sqrt{\frac{(18-19.3)^2+(11-19.3)^2+(22-19.3)^2+(26-19.3)^2}{3}}
\][/tex]
Simplifying inside the square root:
[tex]\[
s=\sqrt{\frac{(18-19.3)^2+(11-19.3)^2+(22-19.3)^2+(26-19.3)^2}{3}}
=s=\sqrt{\frac{(-1.3)^2+(-8.3)^2+(2.7)^2+(6.7)^2}{3}}
\][/tex]
Calculating the squares:
[tex]\[
s=\sqrt{\frac{1.69+68.89+7.29+44.89}{3}}=\sqrt{\frac{122.76}{3}}=\sqrt{40.92}
\][/tex]
Taking the square root:
[tex]\[
s \approx 6.397
\][/tex]
Thus, the standard deviation for the data set is [tex]\( 6.397 \)[/tex].
[tex]\[
s=6.397 \quad \checkmark
\][/tex]
So, the standard deviation is approximately [tex]\( \boxed{6.397} \)[/tex].
