Given:
n = 48, the sample size
m = $25,000, th sample mean
σ = $3000, the maximum variance
The standard deviation is
s = √σ = √(3000) = 54.772
As a rule of thumb, the range is 4 times the standard deviation. Therefore the range is
R = 4*54.772 = 219.09
Half the range is
R/2 = 219.09/2 = 109.45
The required range is
(25000-109.45, 25000+109.45) = (24890.46, 25109.54)
Answer: ($24,890.46, $25,109.54)
Answer:
Step-by-step explanation:
Given that a small manufacturing firm has 250 employees. Fifty have been employed for less than 5 years and 125 have been with the company for over 10 years. So remaining 75 are between 5 and 10 years.
Suppose that one employee is selected at random from a list of the employees
A) Probability that the selected employee has been with the firm less than 5 years = 
B) Probability that the selected employee has been with the firm between 5 and 10 years
= 
C) Probability that the selected employee has been with the firm more than 10 years
= 
a) P(A) = 0.2
P(C) = 0.5
P(A or B) = 0.2+0.3 = 0.5
P(A and C) = 0 (since A and C are disjoint)
Equation would be 45 x 91 / 6-2
Answer would be : -1
Best of luck!
It is 17 bc the 18 times 34