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Attendances at a high school's basketball games were recorded and found to have a sample mean and variance of 490 and 36, respectively. Calculate x ± s,x ± 2s, and x ± 3s.

Answer :

Answer:

a) [tex]x \pm s = 484 \& 496[/tex]

b) [tex]x \pm 2s = 478 \& 502[/tex]

c) [tex]x \pm 3s= 472 \& 508[/tex]

Step-by-step explanation:

sample mean [tex](\overline{x})=490[/tex]

sample variance [tex](\overline{s}^2)=36[/tex]

sample standard deviation [tex](\overline{s}^2)[/tex] will be the square root of the sample variance:

[tex]\overline{s} = \sqrt{36} = 6[/tex]

Now we'll do our calculations:

a) [tex]x \pm s[/tex]

[tex]x \pm s = 490 \pm 6[/tex]

[tex]490 - 6\,\&\, 490 + 6 [/tex]

[tex]484 \& 496[/tex]

b) [tex]x \pm 2s[/tex]

[tex]x \pm 2s = 490 \pm 2(6)[/tex]

[tex]490 - 12\,\&\, 490 + 12[/tex]

[tex]478 \& 502[/tex]

c) [tex]x \pm 3s[/tex]

[tex]x \pm 3s = 490 \pm 3(6)[/tex]

[tex]490 - 18\,\&\, 490 + 18[/tex]

[tex]472 \& 508[/tex]

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