Answer to Question #176748 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

We have that

(a) the mean:

$$\bar x =\frac{\sum fx}{n}=\frac{7376}{20}=368.8$$

(b) the median:

$\frac{n}{2}=\frac{20}{2}=10$ thus class median is the 3rd class

$$median=L_m+i(\frac{\frac{n}{2}-F}{f_m})$$

where

n = the total frequency = 20

F = the cumulative frequency before class median = 9

i = the class width = 8

f_m = the frequency of the class median = 2

L_m= the lower boundary of the class median = 353.5

$median=353.5+8\cdot(\frac{10-9}{2})=357.5$

(c) the variance for sample data is calculated by the formula:

$$s^2=\frac{\sum fx^2-\frac{(\sum fx)^2}{n}}{n-1}=\frac{2743775-\frac{54405376}{20}}{19}=1237.2$$

(d) the standard deviation:

$s=\sqrt {s^2}=\sqrt{1237.2}=35.2$

Answer:

(a) the mean = 368.8

(b) the median = 357.5

(c) the variance for the above frequency intervals = 1237.2

(d) the standard deviation = 35.2