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

(a) Let's find the mean $\overline x = \frac{{\sum {x_i^*{n_i}} }}{n}$ . Build a calculation table:

$\begin{matrix} {interval}&{x_i^*,\,\,middle{\rm{ }}\,of{\rm{ }}\,interval}&{{n_i}}&{x_i^*{n_i}}\\ {318 - 335}&{326.5}&4&{1306}\\ {336 - 353}&{344.5}&5&{1722.5}\\ {354 - 371}&{362.5}&2&{725}\\ {372 - 389}&{380.5}&3&{1141.5}\\ {390 - 407}&{398.5}&2&{797}\\ {408 - 425}&{416.5}&3&{1249.5}\\ {426 - 431}&{434.5}&1&{434.5}\\ {sum}&{}&{20}&{7376} \end{matrix}$

Then $\overline x = \frac{{7376}}{{20}} = {\rm{368}}{\rm{.8}}$

Answer: $\overline x = {\rm{368}}{\rm{.8}}$

(b) Let's make a table of accumulated frequencies ${f_i}$ :

$\begin{matrix} {interval}&{{n_i}}&{{f_i}}&{}\\ {318 - 335}&4&4&{}\\ {336 - 353}&5&9&{}\\ {354 - 371}&2&{11}&{}\\ {372 - 389}&3&{14}&{}\\ {390 - 407}&2&{16}&{}\\ {408 - 425}&3&{19}&{}\\ {426 - 431}&1&{20}&{} \end{matrix}$

Then median interval is 354-371. Then median is

$Me = {x_{Me}} + i\frac{{\frac{{\sum f }}{2} - {f_{Me - 1}}}}{{{f_{Me}}}} = 354 + 17 \cdot \frac{{\frac{{20}}{2} - 9}}{{11}} \approx 355.55$

Answer: $Me \approx 355.55$

(c) Let's find the variance $D = \frac{{\sum {{{\left( {x_i^* - \overline x } \right)}^2}{n_i}} }}{n}$ . Build a calculation table:

$\begin{matrix} {interval}&{x_i^*,\,\,middle\,{\rm{ }}of{\rm{ }} \,interval}&{{n_i}}&{{{\left( {x_i^* - \overline x } \right)}^2}}&{{{\left( {x_i^* - \overline x } \right)}^2}{n_i}}\\ {318 - 335}&{326.5}&4&{1789.29}&{7157.16}\\ {336 - 353}&{344.5}&5&{590.49}&{2952.45}\\ {354 - 371}&{362.5}&2&{39.69}&{79.38}\\ {372 - 389}&{380.5}&3&{136.89}&{410.67}\\ {390 - 407}&{398.5}&2&{882.09}&{1764.18}\\ {408 - 425}&{416.5}&3&{2275.29}&{6825.87}\\ {426 - 431}&{434.5}&1&{4316.49}&{4316.49}\\ {sum}&{}&{20}&{}&{23506.2} \end{matrix}$

Then $D = \frac{{23506.2}}{{20}} = {\rm{1175}}{\rm{.31}}$

Answer: $D = {\rm{1175}}{\rm{.31}}$

(d) Let's find the standard deviation: $\sigma {\rm{ = }}\sqrt D = \sqrt {{\rm{1175}}{\rm{.31}}} \approx 34.28$ .

Answer: $\sigma {\rm{ }} \approx 34.28$