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

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Statistics

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[μx ± 2σx]. (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 [μx ±2σx]. According to the probability distributionof demand x previously given, what percentage of allpossible daily demand values fall in the interval[μx ± 2σx]? (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)

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
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