Question

find the sample variance and standard deviation.
7, 48, 11, 49, 35, 26, 31, 32, 30, 32
choose the correct answer below. fill in the answer box to complete your choice.
(round to two decimal places as needed.)
a. ( s^{2}= )
b. ( sigma^{2}= )

Explanation:

Step1: Calculate the sample mean

The formula for the sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Here \(n = 10\) and \(x=\{7,48,11,49,35,26,31,32,30,32\}\).
\(\sum_{i=1}^{10}x_{i}=7 + 48+11 + 49+35+26+31+32+30+32=291\).
\(\bar{x}=\frac{291}{10}=29.1\).

Step2: Calculate the sum of \((x_{i}-\bar{x})^{2}\)

\((7 - 29.1)^{2}=(- 22.1)^{2}=488.41\)
\((48 - 29.1)^{2}=(18.9)^{2}=357.21\)
\((11 - 29.1)^{2}=(-18.1)^{2}=327.61\)
\((49 - 29.1)^{2}=(19.9)^{2}=396.01\)
\((35 - 29.1)^{2}=(5.9)^{2}=34.81\)
\((26 - 29.1)^{2}=(-3.1)^{2}=9.61\)
\((31 - 29.1)^{2}=(1.9)^{2}=3.61\)
\((32 - 29.1)^{2}=(2.9)^{2}=8.41\)
\((30 - 29.1)^{2}=(0.9)^{2}=0.81\)
\((32 - 29.1)^{2}=(2.9)^{2}=8.41\)
\(\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=488.41+357.21+327.61+396.01+34.81+9.61+3.61+8.41+0.81+8.41 =1634.9\)

Step3: Calculate the sample variance \(s^{2}\)

The formula for the sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\).
Here \(n = 10\), so \(s^{2}=\frac{1634.9}{10 - 1}=\frac{1634.9}{9}\approx181.66\)

Answer:

\(s^{2}=181.66\)