Question #8fc3c

1 Answer
Jun 6, 2017

Answer:

b) #"mean" = 27 2/7, "median" = 29.5, "Mode" = 31, "Range" = 29#

c) #50%# of the dogs weighed less than #30# pounds.

Explanation:

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.