A sample of 100 clients of an exercise facility was selected. Let X = the number of days per week that a randomly selected client uses the exercise facility. X Frequency 0 3 1 15 2 31 3 27 4 10 5 7 6 7 Find the number that is 1.5 standard deviations BELOW the mean. (Round your answer to three decimal places.)

[tex]X\ \ \ \ \ \ \text{Frequency}\\\\ 0\ \ \ \ \ \ 3\\\\ 1 \ \ \ \ \ \ 15 \\\\2 \ \ \ \ \ \ 31 \\\\3 \ \ \ \ \ \ 27 \\\\4 \ \ \ \ \ \ 10 \\\\5 \ \ \ \ \ \ 7 \\\\6 \ \ \ \ \ \ 7[/tex]

standard deviations=1.5

To find:

mean=?

Solution:

[tex]X \ \ \ \ \text{Frequency\ f} \ \ \ \ \ \ fx \ \ \ \ \ \ fx^2\\\\ 0\ \ \ \ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ 0 \\\\ 1 \ \ \ \ \ \ \ \ \ \ \ 15 \ \ \ \ \ \ \ \ \ \ \ 15 \ \ \ \ \ \ \ \ \ \ \ 15\\\\ 2 \ \ \ \ \ \ \ \ \ \ \ 31\ \ \ \ \ \ \ \ \ \ \ 62\ \ \ \ \ \ \ \ \ \ \ 124\\\\ 3\ \ \ \ \ \ \ \ \ \ \ 27 \ \ \ \ \ \ \ \ \ \ \ 81\ \ \ \ \ \ \ \ \ \ \ 243\\\\ 4 \ \ \ \ \ \ \ \ \ \ \ 10\ \ \ \ \ \ \ \ \ \ \ 40 \ \ \ \ \ \ \ \ \ \ \ 160\\\\[/tex]

[tex]5\ \ \ \ \ \ \ \ \ \ \ 7 \ \ \ \ \ \ \ \ \ \ \ 35\ \ \ \ \ \ \ \ \ \ \ 175\\\\ 6\ \ \ \ \ \ \ \ \ \ \ 7\ \ \ \ \ \ \ \ \ \ \ 42\ \ \ \ \ \ \ \ \ \ \ 252 \\\\[/tex]

[tex]100 \ \ \ \ \ \ \ \ \ \ 275 \ \ \ \ \ \ \ \ \ \ 969[/tex]

Calculating the mean:

[tex]\to \bar{x}= \frac{\Sigma fx}{\Sigma f}\\\\[/tex]

[tex]= \frac{275}{100}\\\\= 2.75[/tex]

Calculating the Standard deviation :

[tex]S = \sqrt{\frac{\Sigma fx^2 - \Sigma f \bar{x}^2}{\Sigma f -1}}=\sqrt{\frac{969- 100\times (2.75)^2}{100-1}}\\=\sqrt{\frac{969- 100\times 7.5625}{99}}\\=\sqrt{\frac{969- 756.25}{99}}\\=\sqrt{\frac{212.75}{99}}\\= \sqrt{2.14}\\\\=1.466\\\\Therefore,\\\\\to \bar{x} - 1.5s = 2.75 - 1.5 \times 1.46 = 0.551[/tex]

So, the final answer is "0.551"

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