Answer to Question #175948 in Statistics and Probability for Sean

Question

The monthly income of employees in a small company is summarised in the frequency 

distribution below:

Monthly salary (N$) Number of employees

[1500 – 1999] 15

[2000 – 2499] 10

[2500 – 2999] 20

[3000 – 3499] 4

[3500 – 3999] 1

Required: 

a) Suppose the government regulations stipulate that the minimum monthly wage is N$ 2 500. What 

percentage of employees at this company earn below the stipulated minimum wage? (2)

b) How many employees earn exactly N$ 3 000 per month? (2)

c) How many employees earn at least N$ 3 000 per month? (1)

d) Calculate the following descriptive statistics for the random 

Answer 1

(a) The number of people who earn below minimum age = 15+10=25

(b) The regression coefficient of y on x is-

$b_{yx}=\dfrac{n\sum xy-\sum x \sum y}{n\sum x^2-(\sum x)^2}$

$=\dfrac{5\times 129723.5-120475\times 50}{5\times 40298751.25-(120475)^2}$

$=\dfrac{-38757.5}{12500000}=-0.0031$

The regreesion equation of y on x is-

$y-\bar{y}=b_{yx}(x-\bar{x})$

$y-10=-0.0031(x-2749.5)$

Number of people Who earn exactly $3000

$y=10-0.0031(3000-2749.5)\\\Rightarrow y=10-0.0031(250)=10-0.77=9.22$ ~=9

(c) Number of people who earn at least $3000

=4+1=5

(d) Mean of given data $\mu=\dfrac{\sum xf}{\sum f}=\dfrac{120475}{50}=2409.5$

Variance $\sigma^2 = \dfrac{\sum (x-\mu)^2}{N}=\dfrac{3078000}{50}=61560$

standard deviation $\sigma=\sqrt{variance}=\sqrt{61560}=248.11$