The demand (in number of copies per day) for a city newspaper, x, has historically been...

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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 100,000with the respective probabilities .2, .16, .5, .1, and.04.
(b) Find the expected demand. (Round your answer to thenearest whole number.)
(c)
Using Chebyshev's Theorem, find the minimum percentage of allpossible daily demand values that will fall in the interval[Î¼_x Â± 2Ïƒ_x]. (Roundyour answer to the nearest whole number. Input your answers tominimum percentage and percentage of all possible as percentswithout percent sign.)
(d)
Calculate the interval [Î¼_x Â±2Ïƒ_x]. According to the probability distributionof demand x previously given, what percentage of allpossible daily demand values fall in the interval[Î¼_x Â± 2Ïƒ_x]? (Roundyour intermediate values to the nearest whole number. Round youranswers to the nearest whole number. Input your answers to minimumpercentage and percentage of all possible as percents withoutpercent sign.)

Answer & Explanation Solved by verified expert

3.9 Ratings (500 Votes)

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

Â Â expected demand =65680

from Chebyshev's Theorem ;

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

=75 %

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) =Î¼=	Î£xP(x) =	65680.00
	E(x²) =	Î£x²P(x) =	4483840000.00
	Var(x)=Ïƒ² =	E(x²)-(E(x))²=	169977600.00
	std deviation=Â Â	Â Â Â Â Â Ïƒ= âˆšÏƒ² =	13037.55

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

therefore corresponding probability =P(39604.90

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