Respuesta :
Answer:
(a) Yes, f(x) is a valid probability mass function.
(b) The probability that you will walk at least two dogs this week = 0.74.
(c) Â The expected number of dogs you will walk this week = 3 dogs.
(d) The expected value of X2 = 11.29
(e) Var[x] = E(X2) – (E[X])2 = 3.2811
Step-by-step explanation:
We are given with the probability mass function (pmf),f(x), o'is defined as follows:
Firstly let X = Number of dogs you walk this week
  X         P(X = x)
  0            0.14
  1            0.12
  2            0.15
  3            0.23
  4            0.18
  5            0.09
  6            0.08
  7            0.01
(a) Now f(x) to be a valid probability mass function, two conditions should be   met :
- All values should be >= 0.
- Sum of all probabilities must be equal to 1.
So, First condition is already met as all values are positive and for second condition = 0.14 + 0.12+ 0.15+ 0.23+ 0.18+ 0.09+ 0.08+ 0.01 = 1
Hence both the conditions are satisfied so f(x) is a valid probability mass function.
(b) Probability that we will walk at least two dogs this week = P(X>=2)
   = 1 - P(X = 0) - P(X = 1) = 1 -  0.14 - 0.12 = 0.74
(c) To Compute the expected number of dogs you will walk this week we will use expectation formula which says:
      E(X) = [tex]\frac{\sum X\times P(X=x)}{\sum P(X=x)}[/tex] = [tex]\frac{0*0.14 + 1*0.12 + 2*0.15 + 3*0.23 + 4*0.18 + 5*0.09 + 6*0.08 + 7*0.01}{1}[/tex]
                      = 2.83 or 3 after rounding off.
Therefore the expected number of dogs you will walk this week are 3 dogs.
(d) The expected value of X2 [E(X2)] = Â [tex]\frac{\sum X^{2} \times P(X=x)}{\sum P(X=x)}[/tex]
   = [tex]\frac{0^{2} *0.14 + 1^{2} *0.12 + 2^{2} *0.15 + 3^{2} *0.23 + 4^{2} *0.18 + 5^{2} *0.09 + 6^{2} *0.08 + 7^{2} *0.01}{1}[/tex] = 11.29
(e) Var[x] = E(X2) – (E[X])2
   We have E(X2) = 11.29 and E(X) = 2.83
   Var[x] = [tex]11.29 - (2.83)^{2}[/tex] = 3.2811.
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