Five of your classmates measured the following baseline pulse rates in their worms: 9, 8, 7, 8, and 8 pulses per minute. What is the standard deviation for the measurements?

Respuesta :

Answer:

0.63

Explanation:

Given that:

baseline pulse rates in their worms = 9, 8, 7, 8, and 8 pulses per minute.

What is the standard deviation for the measurements?

Formula for solving standard deviation (σ) = [tex]\sqrt{\frac{E(x_i- x')^2}{N} }[/tex]

where;

N = number of people in the population

[tex]x_i[/tex] = individual entity in the population

x' = population mean

Our data given includes: 9,8,7,8,8 only

re-arranging what we have above in increasing order; we have:

7,8,8,8,9

∴ N = 5

x'(population mean) = [tex]\frac{sum of all values }{total number of thevalue}[/tex]

x'(population mean) = [tex]\frac{7+8+8+8+9}{5}[/tex]

x'(population mean) = [tex]\frac{40}{5}[/tex]

x'(population mean) = 8

     [tex]x_i[/tex]                  ([tex]x_i[/tex]-x')                  ([tex]x_i[/tex]-x')²

     7                   (7-8) = -1              (-1)²=1

     8                   (8-8) = 0             (0)²=0

     8                   (8-8) = 0             (0)²=0

     8                   (8-8) = 0             (0)²=0

    9                   (9-8) = 1               (1)²=1            

[tex]Ex_i[/tex]= 40         [tex]E(x_i-x')[/tex] = 0     [tex]E(x-x')^2[/tex] = 2

(σ) = [tex]\sqrt{\frac{E(x_i- x')^2}{N} }[/tex]

(σ) = [tex]\sqrt{\frac{2}{5} }[/tex]

(σ) = [tex]\sqrt{0.4}[/tex]

(σ) = 0.63248

(σ) = 0.63

∴  the standard deviation for the measurements = 0.63

The standard deviation for the measurements will be "0.63".

Statistics:

According to the question,

Baseline pulse rates: 9, 8, 7, 8, and 8 pulses per minute.

By arranging the terms, we get

7, 8, 8, 8, 9

Here,

Number of people, N = 5

Population mean be:

→ x' = [tex]\frac{Sum \ of \ all \ values}{Total \ number \ of \ values}[/tex]

By substituting the values,

      = [tex]\frac{7+8+8+8+9}{5}[/tex]

      = [tex]\frac{40}{5}[/tex]

      = [tex]8[/tex]

Now,

[tex]x_i[/tex]                                [tex](x_i-x')[/tex]                           [tex](x_i-x')^2[/tex]

7                                7 - 8 = -1                           (-1)² = 1

8                                8 - 8 = 0                          (0)² = 0

8                                8 - 8 = 0                          (0)² = 0

8                                8 - 8 = 0                          (0)² = 0

9                                9 - 8 = 1                            (1)² = 1

[tex]Ex_i=40[/tex]                [tex]E(x_i-x')=0[/tex]                  [tex]E(x-x')=2[/tex]  

hence,

The standard deviation will be:

→ σ = [tex]\sqrt{\frac{E(x_i-x')^2}{N} }[/tex]

By substituting the values,

     = [tex]\sqrt{\frac{2}{5} }[/tex]

     = [tex]\sqrt{0.4}[/tex]

     = [tex]0.63[/tex]

Thus the above response is correct.  

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