Question

Calculate the parameters of the probability distribution below. Interpret the results of your chosen scenario (salary ranges for jobs in Minnesota).

a) Mean

b) Variance

c) Standard Deviation

class class Boundaries Frequency | 40,000. 5- १००००,६ 120 ५०,००) १७ 5०,०००,5-60,०००.5 47 69,000.5-70,000.5 31 170,०००.5-80०००. 5०,००० 5०,००। 6०,००० 6०,००। 10,000 10,००। 80,००० 80,००) १०,००० १०,०० 10०,००० 1००,००० 11०,००० 11०,००। || 20,००० 29 23 80,०००.६-१०,०००.5 १०,०००.5-100,000.5 1०००००.5-110,०००.5 110,०००.5- 120,000.5 १ 10

Answer 1

The table is shown as -

Class Boundaries	Class Mark (x)	Frequency (f)
40000.5-50000.5	45000.5	120
50000.5-60000.5	55000.5	95
60000.5-70000.5	65000.5	47
70000.5-80000.5	75000.5	31
80000.5-90000.5	85000.5	29
90000.5-100000.5	95000.5	23
100000.5-110000.5	105000.5	9
110000.5-120000.5	115000.5	10
Total		364

$\small Mean =\mu=\frac{1}{N}\sum xf, \ where \ N=\sum f$

                     $\small =\frac{1}{364}[(45000.5\times 120)+...+(115000.5\times 10)]=62500.5$

$\small Variance=\sigma^2=\frac{1}{N}\sum (x-\mu)^2f=\frac{1}{N}\sum x^2f-\mu^2$

                                                      = 357211538

$\small Standard \ Deviation=\sigma=\sqrt{357211538}=18900.04$                                 (Answer)