Which statement is true about the graph?
A) A late arrival who is 21 years old with a back-stage pass will make the mean greater than the median.
B) The two holders of back-stage passes whose ages are above 40 make the mean age higher than the median age.
C) The ages of concert-goers with backstage passes are skewed left, so the mean age is less than the median age.
D) A concert-goer who is 18 years old and wins a back-stage pass will pull the mean more than 2 years less than the median.
The two holders of back-stage passes whose ages are above 40 making the mean age higher than the median age is correct about the graph.
This can be defined as pictorial representation of data or values in an organized manner.
From the graph we can calculate mean.
Mean = [13.5(3) + 16.5(6) + 19.5(8) + 22.5(6) + 25.5(5) + 28.5(4) + 31.5(3) + 43.5 + 46.5] / (3 + 6 + 8 + 6 + 5 + 4 + 3 + 1 + 1)
= 856.5 / 37
= 23.15
Median = (37 + 1) / 2 = 38 / 2 = 19th position between 21 - 24 age interval. Median = L1 + C[n/2 - summation (Fl)] / Fm
where L1 = lower class boundary of the median class, C = class size, n = total frequency, summation (Fl) = summation of all frequencies below median class, Fm= median class frequency.
Median = 21 + 4[37/2 - (3 + 6 + 8)] / 6
= 21 + 4[18.5 - 17] / 6
= 21 + 4(1.5) / 6
= 21 + 6 / 6 = 21 + 1 = 22.
Mean of 23.15 is greater than the median of 22.
Mean without two numbers above 40
= (856.5 - 43.5 - 46.5) / 35 = 766.5 / 35 = 21.9.
Therefore the two holders of back-stage passes whose ages are above 40 make the mean age higher than the median age which makes option B the most appropriate choice
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Help me plz
Answer:
Pete worked for 4 hours.
Step-by-step explanation:
If you subtract his tips from his overall total (50-16) you get 34. Now you divide what's left of his total by his hourly wage (34/8.5) which gets you 4.
A. 15
B. 90
C. 20
D. 45
An animal shelter has a ratio of dogs to cats that's 3:2. If there are 30 cats at the shelter, 45 dogs are there . 3/2 = x/30 , x= 30*3/2 =45