Find the median from this data?

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Find the median from this data?

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Answer 1 · 2018-02-11T05:07:17

$150 < t \leq 200$ $150 < t <= 200$

Explanation:

The median is the "middle answer" in a list of answers. For instance, if we have the numbers $1, 2, 6, 7, 9$ $1, 2, 6, 7, 9$ , the median is the middle value, or 6.

For the data above, we can take the frequencies as the number of values we'll be lining up - the middle value will sit in the interval that holds the middle value. Like this:

$10, 8, 24, 29, 32, 10$ $10, 8, 24, 29, 32, 10$

We can add the values and divide by 2:

$\frac{10 + 8 + 24 + 29 + 32 + 10}{2} = 56.5$ $(10+8+24+29+32+10)/2=56.5$

The 56th/57th values sit:

$10 + 8 + 24 = 42$ $10+8+24=42$
$10 + 8 + 24 + 29 = 71 \leftarrow in here$ $10+8+24+29=71 larr "in here"$

And so the interval with the median sits in the $150 < t \leq 200$ $150 < t <= 200$

Answer 2 · 2018-02-11T06:22:15

The median is the $57^{t h}$ $57^(th)$ value which lies in the interval

$150 < t \leq 200$ $150 < t <=200$

Explanation:

If you have an odd number of values, then the median is slightly easier to find than if you have an even number of values.

The median is the middle value in a set of data.

Add the frequencies to find the total number of lightbulbs,

$10 + 8 + 24 + 29 + 32 + 10 = 113$ $10+8+24+29+32+10= 113$

Think of $113$ $113$ as having the same number on each side, with one value in the middle: (that will be the median)

$113 \div 2 = 56 \frac{1}{2}$ $113 div 2 = 56 1/2$

$113 = (56 + \frac{1}{2}) + (\frac{1}{2} + 56)$ $113 = (56+ color(blue)(1/2)) + (color(blue)(1/2) +56)$

$113 = 56 + 1 + 56$ $113 = 56" "+" "color(blue)(1)" "+" "56$

Arrange the given frequencies to look like this.

$10 + 8 + 24 = 42$ $10+8+24 = 42$ , if you now add $29$ $29$ you will have more than $56$ $56$

$10 + 8 + 24 + 29 = 71$ $10+8+24+29=71$

Split up the $29 \to 56 - 42 = 14$ $29 rarr 56-42 = 14$ more needed

$10 + 8 + 24 + 29 + 32 + 10$ $" "10+8+24+color(blue)(29)+32+10$

$10 + 8 + 24 + 14 + 1 + 14 + 32 + 10$ $10+8+24+color(blue)(14+1+14)+32+10$

This gives us what we need:

$(10 + 8 + 24 + 14) + 1 (+ 14 + 32 + 10)$ $color(red)((10+8+24+14))" " color(blue)( +1)" "color(red)((+14+32+10))$
$w w w w w ↓ w w w w w w ↓ w w w w w w w w ↓$ $color(white)(wwwww)darrcolor(white)(wwwwww)darrcolor(white)(wwwwwwww)darr$
$w w w w w 56 w w w w the median w w w w w 56$ $color(white)(wwwww)color(red)(56)color(white)(wwww)color(blue)("the median") color(white)(wwwww)color(red)(56)$

The median therefore lies in the interval which has the frequency of $29$ $29$ because we had to split up $29$ $29$ to get the equal numbers on each side.

This is $150 < t \leq 200$ $150 < t <=200$

Note that the median is the $57^{t h}$ $57^(th)$ value.

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Find the median from this data?

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