Question

The demand (in number of copies per day) for a city newspaper, x, has historically been 46,000, 58,000, 70,000, 82,000, or 100,000 with the respective probabilities .2, .16, .5, .1, and .04.

(b)	Find the expected demand. (Round your answer to the nearest whole number.)

(c)	Using Chebyshev's Theorem, find the minimum percentage of all possible daily demand values that will fall in the interval [μx ± 2σx]. (Round your answer to the nearest whole number. Input your answers to minimum percentage and percentage of all possible as percents without percent sign.)

(d)

Calculate the interval [μx ± 2σx]. According to the probability distribution of demand x previously given, what percentage of all possible daily demand values fall in the interval [μx ± 2σx]? (Round your intermediate values to the nearest whole number. Round your answers to the nearest whole number. Input your answers to minimum percentage and percentage of all possible as percents without percent sign.)

Answer 1

b)

x	P(X=x)	xP(x)
46000	0.200	9200.00000
58000	0.160	9280.00000
70000	0.500	35000.00000
82000	0.100	8200.00000
100000	0.040	4000.00000
total		65680.0000

  expected demand =65680

c)

from Chebyshev's Theorem ;

minimum percentage of all possible daily demand values that will fall in the interva =(1-1/2²)*100

=75 %

d)

x	P(X=x)	xP(x)	x2P(x)
46000	0.200	9200.00000	423200000.00
58000	0.160	9280.00000	538240000.00
70000	0.500	35000.00000	2450000000.00
82000	0.100	8200.00000	672400000.00
100000	0.040	4000.00000	400000000.00
total		65680.0000	4483840000.00
	E(x) =μ=	ΣxP(x) =	65680.00
	E(x2) =	Σx2P(x) =	4483840000.00
	Var(x)=σ2 =	E(x2)-(E(x))2=	169977600.00
	std deviation=	σ= √σ2 =	13037.55

2 standard deviation away values =(65680 -/+2*13037.55 )=39604.90 to 91755.10

therefore corresponding probability =P(39604.90<X<91755.10) =0.2+0.16+0.5+0.1 =0.96~ 96 %