Answer to Question #175751 in Statistics and Probability for kris

Question

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

Answer 1

Construct a table

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

The total number of observations is

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

(a) The mean weight gains is

$\overline{x}=\dfrac{1}{20}\cdot(4\cdot326.5+5\cdot344.5+2\cdot362.5+$

$3\cdot380.5+2\cdot398.5+3\cdot416.5+1\cdot434.5)=$

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

(b) The median

The first interval with cumulative frequence $>\dfrac{n}{2}$ is $[354-371]$ . Previous interval has the frequency $=9$ . The length of any interval is $18$ . The total number of observations is $20$. So the median is

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

(c) The variance

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

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

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

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

(d) The standard deviation

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