Question

Question No. 24 • Analyses of a sample of iron ore gave the following percentage value for iron content: 7.08, 7.21, 7.12, 7.09, 7.16, 7.14, 7.07, 7.14, 7.18, 7.11. Calculate the, standard deviation.​

Answer 1

Answer:

7.08,7.09

Explanation:

tried 7.square

Answer 2

Answer:

The standard deviation of the samples of the iron ores is 0.044.

Explanation:

Given,

Percentages of the iron content:

• 7.08%, 7.21%, 7.12%, 7.09%, 7.16%, 7.14%, 7.07%, 7.14%, 7.18%, 7.11%.

Total number of the iron contents = 10

Sample standard deviation, σ =?

Let the mean = m

Let the terms in the given data = $x_{i}$

As we know,

• The standard deviation is described as the deviation of the given data from an average mean.
• $\sigma =\sqrt{\frac{\sum\limits^n_i ( x_{i}-m)^{2}}{n-1} }$

Firstly, we have to calculate the mean of the iron content:

• $Mean = \frac{sum of the observations}{Total number of the observations}$
• $Mean = \frac{7.08+7.21+7.12+7.09+7.16+7.14+7.07+7.14+7.18+7.11}{10}$
• $Mean = \frac{71.3}{10}$
• Mean = 7.13

Now,

• $x_{i}$          $(x_{i} - m)^{2}$
• 7.08      0.0025
• 7.21       0.0064
• 7.12       0.0001
• 7.09      0.0016
• 7.16       0.0009
• 7.14       0.0001
• 7.07      0.0036
• 7.14       0.0001
• 7.18       0.0025
• 7.11        0.0004

Now, the sum of all the $(x_{i} - m)^{2}$ = 0.0182

After putting the values in the equation of sample standard deviation, we get:

• $\sigma =\sqrt{\frac{0.0182}{10 - 1} }$
• $\sigma =\sqrt{\frac{0.0182}{9} }$
• $\sigma =\sqrt{0.0020}$
• σ = 0.044

Hence, the sample standard deviation, σ = 0.044.