Question #8fc3c

b)

Mean = average of data

Mean =

`(7 +32 + 34 + 31 + 26 + 27 + 23 + 19 + 22 + 29 + 30 + 36 + 35 + 31)/14`

`"mean" = (382)/14`

`color(blue)("mean" = 27 2/7`

Median = middle number, we can find this from listing the numbers from lowest to greatest and finding the middle number.

`7, 32, 34, 31, 26, 27, 23, 19, 22, 29, 30, 36, 35, 31`

`7, 19, 22, 23, 26, 27, 29, 30, 31, 31, 32, 34, 35, 36`

We can take 6 numbers off each side first.

`cancel(7, 19, 22, 23, 26, 27), 29, 30, cancel(31, 31, 32, 34, 35, 36)`

Because there are two medians, we find the mean of them.

`"median" = (29 + 30)/2`

`"median" = 59/2`

`color(blue)("median" = 29.5`

Mode = the number that occurs most often

We can make a tally of each number with how many times they occur in the set of numbers.

`7` - |
`19`- |
`22`- |
`23`- |
`26`- |
`27`- |
`29`- |
`30` - |
`31`- ||
`32` - |
`34` - |
`35` - |
`36` - |

We can see here that above all, `31` occurs twice, when all other numbers occur once.

`color(blue)("Mode" = 31`

Range = the difference between the highest and lowest numbers.

Range = highest number - lowest number

Range = `36 - 7`

`color(blue)("Range" = 29`

c)

we can see above that out of the `14` dogs, `7` weighed less than `30` pounds. We can also say that a percentage is a number over `100`, or `x/100`

So to find the percentage, we can make this equation:

`7/14 = x/100`

Now we can find out how the first denominator gets to the next, so we can do the same to the numerator. We can make this formula to find `x`.

`(7 xx y = x)/(14 xx y = 100)`

`14 xx y = 100`

`y = 100 ÷ 14`

`y ~~ 7.147`

We can now put `7.147` in the equation in substitution for `y`

`(7 xx 7.147 = x)/(14 xx 7.147 = 100)`

Now we can find `x`.

`x = 7 xx 7.147`

`x = 50`

`x = 50%`

`therefore` `50%` of the dogs weighed less than `30` pounds.