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Sagot :
1) First, from the question we see that we have a table with the probability distribution p(X) for the number of cheesecakes (X) sold on a randomly selected day. We know that the numbers in the table for P(X) should sum up to 1, that's because the total probability always sums 1. So using this fact we can see that:
[tex]P(x=15)=0.28[/tex]2) The probability of selling at least 10 cheesecakes is the sum of probabilities P(x) for x ≥ 10, using the data from the table and the probability obtained above we have:
[tex]\begin{gathered} P(x\ge10)=P(x=10)+P(x=15)+P(x=20) \\ P(x\ge10)=0.21+0.28+0.1 \\ P(x\ge10)=0.59 \end{gathered}[/tex]3) The probability of selling 5 or 15 cheesecakes is the joint probability of the events of selling 5 cheesecakes P(x = 5) or 15 cheesecakes P(x = 15) because they are independent events (i.e. P(x=5 ∩ x=15) = 0), we have:
[tex]\begin{gathered} P(x=5orx=15)=P(x=5)+P(x=15)-P(x=5andx=15) \\ P(x=5orx=15)=0.3+0.28-0 \\ P(x=5orx=15)=0.58 \end{gathered}[/tex]4) From the table we see that we don't have an assigned value for the probability of selling x = 25 cheesecakes, so the probability for this event is zero:
[tex]P(x=25)=0[/tex]5) The probability of selling at most 10 cheesecakes is the sum of the probabilities P(x) for x ≤ 10, using the data from the table we have:
[tex]\begin{gathered} P(x\leq10)=P(x=0)+P(x=5)+P(x=10) \\ P(x\leq10)=0.11+0.3+0.21 \\ P(x\leq10)=0.62 \end{gathered}[/tex]6) Finally, we must compute the expected value μ of cheesecakes sold on any given day, applying the following formula and the data of the table we get:
[tex]\begin{gathered} \mu=\sum ^{}_iX_i\cdot P(X_i) \\ \mu=0\cdot0.11+5\cdot0.3+10\cdot0.21+15\cdot0.28+20\cdot0.1 \\ \mu=9.8 \end{gathered}[/tex]Answers summary:
1) P(x = 15) = 0.28
2) P(x ≥ 10) = 0.59
3) P(x = 5 or x = 15) = 0.58
4) P(x = 25) = 0
5) P(x ≤ 10) = 0.62
6) μ = 9.8
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