Question
In: Statistics and Probability
The demand (in number of copies per day) for a city newspaper, x, has historically been...
|
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.) |
Solutions
Expert Solution
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 %
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