Respuesta :
Answer:
a) [tex] P(X\geq 1) = 1-P(X<1) = 1-P(X=0) =1-0.01= 0.99[/tex]
b) [tex] P(X\geq 5) = 1-P(X<5) = 1-P(X\leq 4) = 1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)][/tex]
And replacing we got:
[tex] P(X \geq 5) = 1-[0.01+0.02+0.03+0.12+0.11]=1-0.29=0.71[/tex]
Step-by-step explanation:
For this case we can solve this problem creating the following table
Number of particles    Frequency     Rel. Frequency
       0               1            1/100 =0.01
       1               2            2/100 =0.02
       2               3            3/100=0.03
       3               12           12/100=0.12  Â
       4               11            11/100=0.11
       5               15            15/100=0.15
       6               18            18/100=0.18
       7               10            10/100=0.1
       8               12            12/100=0.12
       9               4             4/100=0.04
       10              5             5/100=0.05
       11               3             3/100=0.03
       12              1              1/100=0.01
       13              2             2/100=0.02
       14              1              1/100=0.01
      Total            100               1
We assume on this case the the relative frequency represent the probability.
Let X the number of contaminating particles on a silicon wafer
What proportion of the sampled wafers had at least one particle?
For this case we can use the complement rule like this:
[tex] P(X\geq 1) = 1-P(X<1) = 1-P(X=0) =1-0.01= 0.99[/tex]
At least five particles?
Again for this case we can use the complement rule and we got:
[tex] P(X\geq 5) = 1-P(X<5) = 1-P(X\leq 4) = 1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)][/tex]
And replacing we got:
[tex] P(X \geq 5) = 1-[0.01+0.02+0.03+0.12+0.11]=1-0.29=0.71[/tex]
