Question
Find the mean and standard deviation for the following data set. Round to the nearest tenth.
Data set: 11, 14, 15, 15, 16, 16, 17, 18, 19, 23
A) 16.6 is a typical value for the data set. 3.7 is the typical distance that a value in the data set is from 16.6.
B) 16.4 is a typical value for the data set. 3.0 is the typical distance that a value in the data set is from 16.4.
C) 16.6 is a typical value for the data set. 3.7 is the typical distance between any pair of values in the data set.
D) 16.4 is a typical value for the data set. 3.0 is the typical distance between any pair of values in the data set.
Data set: 11, 14, 15, 15, 16, 16, 17, 18, 19, 23
A) 16.6 is a typical value for the data set. 3.7 is the typical distance that a value in the data set is from 16.6.
B) 16.4 is a typical value for the data set. 3.0 is the typical distance that a value in the data set is from 16.4.
C) 16.6 is a typical value for the data set. 3.7 is the typical distance between any pair of values in the data set.
D) 16.4 is a typical value for the data set. 3.0 is the typical distance between any pair of values in the data set.
Answers
GPT-4o mini
To find the mean and standard deviation for the provided data set: 11, 14, 15, 15, 16, 16, 17, 18, 19, 23, we will follow these steps:
**1. Calculate the Mean:**
The formula for the mean (average) is:
\[
\text{Mean} = \frac{\sum \text{data}}{n}
\]
Where \( n \) is the number of data points.
Calculating the sum:
\[
11 + 14 + 15 + 15 + 16 + 16 + 17 + 18 + 19 + 23 = 11 + 14 + 15 + 15 + 16 + 16 + 17 + 18 + 19 + 23 = 14 + 15 + 15 + 16 + 16 + 17 + 18 + 19 + 23 = 11 + 14 + 15 + 15 + 16 + 16 + 17 + 18 + 19 + 23 = 165
\]
Now dividing by the number of values (10):
\[
\text{Mean} = \frac{165}{10} = 16.5
\]
**2. Calculate the Standard Deviation:**
The formula for the standard deviation \( s \) of a sample is:
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
\]
Where \( \bar{x} \) is the mean calculated above and \( x_i \) represents each value in the data set.
First, we find each \( (x_i - \bar{x})^2 \):
\[
\begin{align*}
(11 - 16.5)^2 & = ( -5.5 )^2 = 30.25 \\
(14 - 16.5)^2 & = ( -2.5 )^2 = 6.25 \\
(15 - 16.5)^2 & = ( -1.5 )^2 = 2.25 \\
(15 - 16.5)^2 & = ( -1.5 )^2 = 2.25 \\
(16 - 16.5)^2 & = ( -0.5 )^2 = 0.25 \\
(16 - 16.5)^2 & = ( -0.5 )^2 = 0.25 \\
(17 - 16.5)^2 & = (0.5)^2 = 0.25 \\
(18 - 16.5)^2 & = (1.5)^2 = 2.25 \\
(19 - 16.5)^2 & = (2.5)^2 = 6.25 \\
(23 - 16.5)^2 & = (6.5)^2 = 42.25 \\
\end{align*}
\]
Now sum these results:
\[
30.25 + 6.25 + 2.25 + 2.25 + 0.25 + 0.25 + 0.25 + 2.25 + 6.25 + 42.25 = 92.5
\]
Now we calculate the standard deviation:
\[
s = \sqrt{\frac{92.5}{10 - 1}} = \sqrt{\frac{92.5}{9}} \approx \sqrt{10.28} \approx 3.2
\]
Finally, we round our mean and standard deviation to the nearest tenth, we get:
- Mean \( \approx 16.5 \)
- Standard Deviation \( \approx 3.2 \)
Comparing with the options given:
- A) 16.6 is a typical value for the data set. 3.7 is the typical distance that a value in the data set is from 16.6.
- B) 16.4 is a typical value for the data set. 3.0 is the typical distance that a value in the data set is from 16.4.
- C) 16.6 is a typical value for the data set. 3.7 is the typical distance between any pair of values in the data set.
- D) 16.4 is a typical value for the data set. 3.0 is the typical distance between any pair of values in the data set.
The closest rounded match is **A)** with mean approximately 16.6 (instead of 16.5, but close) and is also the distance from individual values, while our stdev calculation was near 3.2 (not matching exactly).
Therefore, the answer closest to our calculation is **A)**.
**1. Calculate the Mean:**
The formula for the mean (average) is:
\[
\text{Mean} = \frac{\sum \text{data}}{n}
\]
Where \( n \) is the number of data points.
Calculating the sum:
\[
11 + 14 + 15 + 15 + 16 + 16 + 17 + 18 + 19 + 23 = 11 + 14 + 15 + 15 + 16 + 16 + 17 + 18 + 19 + 23 = 14 + 15 + 15 + 16 + 16 + 17 + 18 + 19 + 23 = 11 + 14 + 15 + 15 + 16 + 16 + 17 + 18 + 19 + 23 = 165
\]
Now dividing by the number of values (10):
\[
\text{Mean} = \frac{165}{10} = 16.5
\]
**2. Calculate the Standard Deviation:**
The formula for the standard deviation \( s \) of a sample is:
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
\]
Where \( \bar{x} \) is the mean calculated above and \( x_i \) represents each value in the data set.
First, we find each \( (x_i - \bar{x})^2 \):
\[
\begin{align*}
(11 - 16.5)^2 & = ( -5.5 )^2 = 30.25 \\
(14 - 16.5)^2 & = ( -2.5 )^2 = 6.25 \\
(15 - 16.5)^2 & = ( -1.5 )^2 = 2.25 \\
(15 - 16.5)^2 & = ( -1.5 )^2 = 2.25 \\
(16 - 16.5)^2 & = ( -0.5 )^2 = 0.25 \\
(16 - 16.5)^2 & = ( -0.5 )^2 = 0.25 \\
(17 - 16.5)^2 & = (0.5)^2 = 0.25 \\
(18 - 16.5)^2 & = (1.5)^2 = 2.25 \\
(19 - 16.5)^2 & = (2.5)^2 = 6.25 \\
(23 - 16.5)^2 & = (6.5)^2 = 42.25 \\
\end{align*}
\]
Now sum these results:
\[
30.25 + 6.25 + 2.25 + 2.25 + 0.25 + 0.25 + 0.25 + 2.25 + 6.25 + 42.25 = 92.5
\]
Now we calculate the standard deviation:
\[
s = \sqrt{\frac{92.5}{10 - 1}} = \sqrt{\frac{92.5}{9}} \approx \sqrt{10.28} \approx 3.2
\]
Finally, we round our mean and standard deviation to the nearest tenth, we get:
- Mean \( \approx 16.5 \)
- Standard Deviation \( \approx 3.2 \)
Comparing with the options given:
- A) 16.6 is a typical value for the data set. 3.7 is the typical distance that a value in the data set is from 16.6.
- B) 16.4 is a typical value for the data set. 3.0 is the typical distance that a value in the data set is from 16.4.
- C) 16.6 is a typical value for the data set. 3.7 is the typical distance between any pair of values in the data set.
- D) 16.4 is a typical value for the data set. 3.0 is the typical distance between any pair of values in the data set.
The closest rounded match is **A)** with mean approximately 16.6 (instead of 16.5, but close) and is also the distance from individual values, while our stdev calculation was near 3.2 (not matching exactly).
Therefore, the answer closest to our calculation is **A)**.