Question

The following random sample of annual salaries was recorded. Units are in thousands

45 50 40 42 30 51 46 10 52 47

a) Calculate the mean and the standard deviation.

b) Using Chebyshev's between what two bounds will at least 80% of the data lie?

c) Using Chebyshev's between what two bounds will at least 95% of the data values lie?

Answer 1

Ans:

a)

	xi	(xi-41.3)^2
1	45	13.69
2	50	75.69
3	40	1.69
4	42	0.49
5	30	127.69
6	51	94.09
7	46	22.09
8	10	979.69
9	52	114.49
10	47	32.49
Total	413	1462.1
mean	41.3

mean=413/10=41.3

standard deviation=sqrt(1462.1/10)=12.1

b)

k=1/sqrt(1-0.8)=2.236

lower bound=41.3-2.236*12.1=14.24

upper bound=41.3+2.236*12.1=68.36

c)

k=1/sqrt(1-0.95)=4.472

lower bound=41.3-4.472*12.1=-12.81, it is negative,as salary can not be negative,consider it 0.

upper bound=41.3+4.472*12.1=95.41