gasoline wholesale distributor has bulk storage tanks holding a fixed supply. The tanks are filled every Monday. Of interest to
1 answer:
Answer:
e. 0.08
Step-by-step explanation:
In the question above, a certain quantity of goods was supplied while a specific quantity of goods was sold per week. In a given week, if the number of proportion sold is X, therefore:
f(x) = {Γ(4+2)/Γ(4)Γ(2) x^3 (1-x), 0≤x≤1 ; 0, elsewhere
and
P(X greater than 0.9) = = 20*{(y^4/4)[1,0.9] - (y^5/5)[1,0.9]} = 20*{(0.25 - 0.164) - (0.20 - 0.118)} = 20*{0.086 - 0.0819} = 20*0.0041 = 0.082
Therefore the probability of the proportion sold is approximately 0.082
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<h2>Answer:</h2><h2>The probability of selecting a defective roll followed by another defective roll = 
</h2>
Step-by-step explanation:
The total number of rolls in a box = 15
The number of defective items = 3
Two rolls are to be selected, one after the other. The probability of selecting a defective roll followed by another defective roll = ?
The probability that the first roll is detective =
The probability that the second roll is detective =
Required probability =
=
Answer:
33.33% probability that it takes Isabella more than 11 minutes to wait for the bus
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The probability that we find a value X lower than x is given by the following formula.
For this problem, we have that:
Uniformly distributed between 3 minutes and 15 minutes:
So
What is the probability that it takes Isabella more than 11 minutes to wait for the bus?
Either she has to wait 11 or less minutes for the bus, or she has to wait more than 11 minutes. The sum of these probabilities is 1. So
We want P(X > 11). So
33.33% probability that it takes Isabella more than 11 minutes to wait for the bus