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3 years ago

13

Ok i forgot how to do this.

1 answer:

d1i1m1o1n [39]3 years ago

3 0

Well, the first dot is the known as the Minimum. Then, the second dot is the 1st Quartile. The dot in the middle of the box with the line through it is the median (2nd Quartile). Then the last dot forming the box is the 3rd Quartile. And the final dot is the 4th Quartile (Also known as the Maximum). In this case, it is asking about the Median...which would be 15.5. Hope this helped!

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Yanka [14]

Answer:

$Range=14$

$\sigma^2 =32.4$

$\sigma = 5 .7$

The standard deviation will remain unchanged.

Step-by-step explanation:

Given

$Data: 136, 129, 141, 139, 138, 127$

Solving (a): The range

This is calculated as:

$Range = Highest - Least$

Where:

$Highest = 141; Least = 127$

So:

$Range=141-127$

$Range=14$

Solving (b): The variance

First, we calculate the mean

$\bar x = \frac{1}{n} \sum x$

$\bar x = \frac{1}{6} (136+ 129+ 141+ 139+ 138+ 127)$

$\bar x = \frac{1}{6} *810$

$\bar x = 135$

The variance is calculated as:

$\sigma^2 =\frac{1}{n-1}\sum(x - \bar x)^2$

So, we have:

$\sigma^2 =\frac{1}{6-1}*[(136 - 135)^2 +(129 - 135)^2 +(141 - 135)^2 +(139 - 135)^2 +(138 - 135)^2 +(127 - 135)^2]$

$\sigma^2 =\frac{1}{5}*[162]$

$\sigma^2 =32.4$

Solving (c): The standard deviation

This is calculated as:

$\sigma = \sqrt {\sigma^2 }$

$\sigma = \sqrt {32.4}$

$\sigma = 5 .7$ --- approximately

Solving (d): With the stated condition, the standard deviation will remain unchanged.

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6×100
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