
a.
Find the
a.

Answer to Problem 13P
The range is 15.
Explanation of Solution
The range is obtained below:
Range=Largest value−Smallest value=30−15=15
Thus, the range is 15.
b.
Verify that ∑x=110 and ∑x2=2568 using calculator.
b.

Explanation of Solution
Step-by-step procedure to verify ∑x=110 and ∑x2=2568 using Ti83 calculator:
- Press STAT.
- Select Edit.
- Enter the values in L1.
- Press STAT and Choose CALC.
- Select 1-Var Stats.
- To select the variable L1, Press 2-nd and then press 1.
- Press Enter.
Output obtained using the Ti83 calculator is given below:
From the output, the values of ∑x=110 and ∑x2=2,568.
c.
Find the sample variance and the sample standard deviation using the computation formula.
c.

Answer to Problem 13P
The sample variance using the computation formula is 37.
The sample standard deviation using the computation formula is 6.08.
Explanation of Solution
The sample variance using the computation formula is given below:
s2=∑x2−(∑x)2nn−1
Where ˉx is the mean, n is the number of values in the data, and x is the value in the data.
The sample standard deviation using the computation formula is given below:
s=√∑x2−(∑x)2nn−1
Where ˉx is the mean, n is the number of values in the data, and x is the value in the data.
The sample variance using the computation formula is obtained below:
s2=2,568−(110)255−1=37
Thus, the sample variance using the computation formula is 37.
The sample standard deviation using the computation formula is obtained below:
s=√2,568−(110)255−1=√37=6.08
Thus, the sample standard deviation using the computation formula is 6.08.
d.
Find the sample variance using the defining formula.
Find the sample standard deviation using the defining formula.
d.

Answer to Problem 13P
The sample variance using the defining formula is 37.
The sample standard deviation using the defining formula is 6.08.
Explanation of Solution
The sample variance using the defining formula is given below:
s2=∑(x−ˉx)2n−1
Where ˉx is the mean, n is the number of values in the data, and x is the value in the data.
The sample standard deviation using the defining formula is given below:
s=√∑(x−ˉx)2n−1
Where ˉx is the mean, n is the number of values in the data, and x is the value in the data.
The value of ˉx is obtained below:
ˉx=∑xn=23+17+15+30+255=22
The value of ∑(x−ˉx)2 is obtained below:
x | x−ˉx | (x−ˉx)2 |
23 | 1 | 1 |
17 | –5 | 25 |
15 | –7 | 49 |
30 | 8 | 64 |
25 | 3 | 9 |
∑(x−ˉx)2=148 |
The sample variance using the defining formula is given below:
s2=1485−1=1484=37
Thus, the sample variance using the defining formula is 37.
The sample standard deviation using the defining formula is given below:
s=√1485−1=√1484=6.08
Thus, the sample standard deviation using the defining formula is 6.08.
e.
Find the population variance using the defining formula.
Find the population standard deviation using the defining formula.
e.

Answer to Problem 13P
The population variance using the defining formula is 29.6.
The population standard deviation using the defining formula is 5.44.
Explanation of Solution
The population variance using the defining formula is given below:
σ2=∑(x−μ)2n
Where μ is the population mean, N is the number of values in the population, and x is the value in the data.
The population standard deviation using the defining formula is given below:
σ=√∑(x−μ)2N
Where μ is the population mean, N is the number of values in the population, and x is the value in the data.
The value of μ is obtained below:
μ=∑xN=23+17+15+30+255=22
The value of ∑(x−μ)2 is obtained below:
x | x−μ | (x−μ)2 |
23 | 1 | 1 |
17 | –5 | 25 |
15 | –7 | 49 |
30 | 8 | 64 |
25 | 3 | 9 |
∑(x−μ)2=148 |
The population variance using the defining formula is given below:
σ2=1485=29.6
Thus, the population variance using the defining formula is 29.6.
The population standard deviation using the defining formula is given below:
σ=√1485=5.44
Thus, the population standard deviation using the defining formula is 5.44.
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