Answer to Question #204071 in Statistics and Probability for Shazz

Question #204071

The age of the population in Wakanda is shown in the following table in the form of a grouped frequency distribution.

  Age (years)

  Number (‘000)

 0 Under 5

5 under 15 15 under 30 30 under 50 50 under 65 65 under 80 80 under 100

Total

Use the data in the table above to estimate for of population in Wakanda the:

1 375.3 2 750.5 4 514.7 6 137.7 3 821.7 2 045.7 783.2

        100 under 110

what is the modal age?



1
Expert's answer
2021-06-08T14:56:55-0400

The age of the population in Wakanda is shown in the following table in the form of a

grouped frequency distribution


"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c}\n Age (years) & Number('000) & Cumulative\\\\ \n& & frequency\\\\ \\hline\n 0-5 & 1375.3 & 1375.3 \\\\\n 5-15 & 2750.5 & 4125.8 \\\\\n 15-30 & 4514.7& 8640.5 \\\\\n30-50 & 6137.7 & 14778.2 \\\\\n 50-65 & 3821.7 & 18559.9 \\\\\n65-80 & 2045.7 & 20645.6 \\\\\n80-100 & 783.2 & 21428.8 \\\\\n100-110 & 3.1 & 21431.9 \\\\\n\\hline \nTotal & 21431.9\n\\end{array}"

Median number "(m)=n\/2=21431.9\/2=10715.95"

m = 10715.95

So, median class = 30 - 50


Median age 

"M=l_1+\\dfrac{(n\/2-cf)}{f}\\cdot l"


where "l_1" = lower number of median class

n/2=median number

cf = cumulative frequency of class preceding to median class

f = frequency/number of median class

i = class length

and we have

"l_1" = 30

n/2 = 10715.95

cf = 8640.5

f = 6137.7

i = 20

Therefore, Median 

"M=l_1+\\dfrac{(n\/2-cf)}{f}\\cdot l"




"=30+[\\dfrac{10715.95-8640.5}{6137.7}]\\cdot 20"




"=30+6.76"

"=36.76"

Hence, median age = 36.76 years


The modal class is "30-50."


"Mode=l_m+(\\dfrac{\\Delta_1}{\\Delta_1+\\Delta_2})C"

"l_m=29.5, \\Delta_1=6137.7-4514.7=1623"

"\\Delta_1=6137.7-3821.7=2316"

"C=50.5-29.5=21"

"Mode=29.5+(\\dfrac{1623}{1623+2316})21=38.15"

Hence, modal age = 38.15 years




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