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[tex]n = 20[/tex] , ∑[tex]x = 1570[/tex] , ∑[tex]x^2 = 125696[/tex] , minimum = 60, maximum = 100

Find s. Round your answer to one decimal place.

Estimate s by using the range rule-of-thumb. (If you can explain the range rule of thumb it'll be greatly appreciated).

Sagot :

Answer:

s=10

Step-by-step explanation:

Basically, the range rule-of-thumb is that the range is generally about four times the standard deviation.

First, we find the range:

Range=Maximum-Minimum

Range=100-60

Range=40

Next, divide the range by 4 to get the standard deviation

Range=4(Standard Deviation)

40=4(Standard Deviation)

10=Standard Deviation

Therefore, s=10, which is the standard deviation

Mean:-

[tex]\\ \tt\hookrightarrow \overline{x}=\dfrac{\sum x}{n}=\dfrac{1570}{20}=78.5[/tex]

Standard deviation:-

  • Two ways are available ,Let's do in both.

WAY-1(apply common formula)

[tex]\\ \tt\hookrightarrow \sigma=\sqrt{\dfrac{\sum x^2}{n}-(\overline{x})^2}[/tex]

[tex]\\ \tt\hookrightarrow \sigma=\sqrt{\dfrac{125696}{20}-(78.5)^2}[/tex]

[tex]\\ \tt\hookrightarrow \sigma=\sqrt{6284.8-6162.25}[/tex]

[tex]\\ \tt\hookrightarrow \sigma=\sqrt{122.55}[/tex]

[tex]\\ \tt\hookrightarrow \sigma=11.07=11.1[/tex]

Way-2(Apply range-rule of thumb):-

  • So the rule basically defines that range is approximately equal to four times of standard deviation.

[tex]\\ \tt\hookrightarrow Range=max-min=100-60=40[/tex]

NOW

[tex]\\ \tt\hookrightarrow \sigma \approx \dfrac{Range}{4}[/tex]

[tex]\\ \tt\hookrightarrow \sigma\approx \dfrac{40}{4}[/tex]

[tex]\\ \tt\hookrightarrow \sigma\approx 10[/tex]

  • Make sure to use approx sign instead of equal to as range -rule of thumb doesn't give accurate standard deviation.
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