Question #175751

Plant scientists developed different varieties of corns that have a rich content of lysine which is a nutritious animal feed. A group of chicks were given this food to test the quality. The distribution of the weight gains (in grams) of these chicks are shown below:


Weight gains (in grams) Frequency

318 - 335 4

336 - 353 5

354 - 371 2

372 - 389 3

390 – 407 2

408 – 425 3

426 - 443 1


Find:

(a) the mean weight gains

(b) the median

(c) the variance for the above frequency intervals

(d) the standard deviation


1
Expert's answer
2021-03-30T08:05:31-0400

Construct a table

IntervalsFrequenciesMidpoints of intervalsCumulative frequencies3183354326.543363535344.593543712362.5113723893380.5143904072398.5164084253416.5194264431434.520\begin{array}{|c|c|c|c|}\hline \rm{Intervals} & \rm{Frequencies} & \rm{Midpoints\ of\ intervals}&\rm{Cumulative\ frequencies}\\ \hline 318 - 335& 4& 326.5& 4\\ \hline 336 - 353& 5& 344.5& 9\\ \hline 354 - 371& 2& 362.5& 11\\ \hline 372 - 389& 3& 380.5& 14\\ \hline 390 - 407& 2& 398.5& 16\\ \hline 408 - 425& 3& 416.5& 19\\ \hline 426 - 443& 1& 434.5& 20\\ \hline \end{array}


The total number of observations is

n=4+5+2+3+2+3+1=20n=4+5+2+3+2+3+1=20


(a) The mean weight gains is


x=120(4326.5+5344.5+2362.5+\overline{x}=\dfrac{1}{20}\cdot(4\cdot326.5+5\cdot344.5+2\cdot362.5+


3380.5+2398.5+3416.5+1434.5)=3\cdot380.5+2\cdot398.5+3\cdot416.5+1\cdot434.5)=

1207376=368.8\dfrac{1}{20}\cdot7376=368.8



(b) The median

The first interval with cumulative frequence >n2>\dfrac{n}{2} is [354371][354-371] . Previous interval has the frequency =9=9 . The length of any interval is 1818 . The total number of observations is 2020. So the median is

Me=354+18122092=363Me=354+18\cdot\dfrac{\frac12\cdot20-9}{2}=363


(c) The variance


Var[X]=x2x2=\bold{Var}[X]=\overline{x^2}-\overline{x}^2=


120(4(326.5)2+5(344.5)2+2(362.5)2+\dfrac{1}{20}\cdot(4\cdot(326.5)^2+5\cdot(344.5)^2+2\cdot(362.5)^2+


3(380.5)+2(398.5)2+3(416.5)2+1(434.5)2)3\cdot(380.5)^+2\cdot(398.5)^2+3\cdot(416.5)^2+1\cdot(434.5)^2)-


(368.8)2=(368.8)^2= 137,188.75136,013.44=1,175.31137,188.75-136,013.44=1,175.31



(d) The standard deviation


s=Var[X]=1,175.3134.28s=\sqrt{\bold{Var}[X]}=\sqrt{1,175.31}\approx34.28




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