Answer:
a) Mode number = 2
b) Median = 64.5 = 4 people in car as 0.5 in 64.5 is close to 76
= 4 people
c) Mean number is 505/6 = 84.1666666667 then we see where 84.16 lies and see this lies with = 3 people
Step-by-step explanation:
45 is 1
198 is 2
121 is 3
76 is 4
52 is 5
13 is 6
Mode we look for the highest frequency and see 198 is the amount and 2 people in car is the subject, we give the subject and that is 2 people in car is the highest frequency and would be the mode.
Median is shown in the answer as we add up.
Mean we add up all frequency and divide by the amount of subjects = 6.
a) Mode number = 2
b) Median = 13, 45, 52, here 76, 121, 198 mid number between
= (76-52 )/ 2 + (52)= 24 /2 + (52) = 12 + 52 + 0.5 if even number
median = 64.5 in a cumulative graph though which is not asked here it would be exactly half of 505 = 257.5 and interquartile we half again and add on 257.5 and show the range values interquartile as 1/2(252.5) and 1/2(252.5+ 505) = 126.25 as one value and 757.5 as the other by drawing a horizontal line from y axis to the points so vertical lines can represnt the cars median and interquartile.
c. Mean number is 505/6 = 84.1666666667 then we see where 84.16 lies and see this lies with 3 seats
For cumulative frequency we can relist numbers like this
45
45 +198 = 243
243 + 121 = 364
364+76 = 440
440 + 52 = 492
492+13 = 505 to plot graph.
But if there was group of measures or time etc then we would always plot the highest of each set for cumulative x axis and use the 2 values to see the ratio for histograms ie) 198/2 = 99 so that all values are read in equal proportions and 2 is a data below the graph not on the graph as 2 does not show as a box count just a name below on x axis. So we have to use the ratio as explained for histograms.
means that from statement q implies statement p.