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n200080 [17]
3 years ago
14

Mum bought a litre of orange juice. She filled 4 glasses and had 1/2 litre left. How much was in each glass?

Mathematics
1 answer:
Bumek [7]3 years ago
3 0
There was 1/8 liter of orange juice in each glass
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A store selling newspapers orders only n = 4 of a certain newspaper because the manager does not get many calls for that publica
umka2103 [35]

Answer:

a) The expected value is 2.680642

b) The minimun number of newspapers the manager should order is 6.

Step-by-step explanation:

a) Lets call X the demanded amount of newspapers demanded, and Y the amount of newspapers sold. Note that 4 newspapers are sold when at least four newspaper are demanded, but it can be <em>more</em> than that.

X is a random variable of Poisson distribution with mean \mu = 3 , and Y is a random variable with range {0, 1, 2, 3, 4}, with the following values

  • PY(k) = PX(k) = ε^(-3)*(3^k)/k! for k in {0,1,2,3}
  • PY(4) = 1 -PX(0) - PX(1) - PX(2) - PX (3)

we obtain:

PY(0) = ε^(-3) = 0.04978..

PY(1) = ε^(-3)*3^1/1! = 3*ε^(-3) = 0.14936

PY(2) = ε^(-3)*3^2/2! = 4.5*ε^(-3) = 0.22404

PY(3) = ε^(-3)*3^3/3! = 4.5*ε^(-3) = 0.22404

PY(4) = 1- (ε^(-3)*(1+3+4.5+4.5)) = 0.352768

E(Y) = 0*PY(0)+1*PY(1)+2*PY(2)+3*PY(3)+4*PY(4) =  0.14936 + 2*0.22404 + 3*0.22404+4*0.352768 = 2.680642

The store is <em>expected</em> to sell 2.680642 newspapers

b) The minimun number can be obtained by applying the cummulative distribution function of X until it reaches a value higher than 0.95. If we order that many newspapers, the probability to have a number of requests not higher than that value is more 0.95, therefore the probability to have more than that amount will be less than 0.05

we know that FX(3) = PX(0)+PX(1)+PX(2)+PX(3) = 0.04978+0.14936+0.22404+0.22404 = 0.647231

FX(4) = FX(3) + PX(4) = 0.647231+ε^(-3)*3^4/4! = 0.815262

FX(5) = 0.815262+ε^(-3)*3^5/5! = 0.91608

FX(6) = 0.91608+ε^(-3)*3^6/6! = 0.966489

So, if we ask for 6 newspapers, the probability of receiving at least 6 calls is 0.966489, and the probability to receive more calls than available newspapers will be less than 0.05.

I hope this helped you!

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