a)
The table shows the values of the function P(X) for every value of X from 0 to 10; then, as suggested by the question, to determine the probability that at least 5 users have abandoned their landlines, add P(5) to P(10), as shown below
[tex]P(X\ge5)=P(5)+P(6)+...+P(10)=0.2087+0.1209+...+0.01=0.402[/tex]Alternatively,
[tex]\begin{gathered} P(X\ge5)=1-P(X<5)=1-(P(0)+P(1)+P(2)+P(3)+P(4)) \\ =1-(0.6078) \\ =0.3922 \end{gathered}[/tex]There is a difference between both results because the data in the table add up to 1.0098, not 1 as it should be.
b)
The binomial distribution states that
[tex]\begin{gathered} P(X=k)=(nbinomialk)p^k(1-p)^{n-k} \\ where \\ (nbinomialk)=\frac{n!}{(n-k)!k!},n!=1*2*3*...*n \\ n\rightarrow\text{ total number of trials} \\ k\rightarrow\text{ number of successful trials} \\ p\rightarrow\text{ probability of a successful trial} \end{gathered}[/tex]In our case, consider that a successful trial consists of a user abandoning their landline; thus,
[tex]\begin{gathered} n=10,k=5,p=44\%=\frac{44}{100}=0.44 \\ \end{gathered}[/tex]Thus, the corresponding probability is
[tex]\begin{gathered} P(X=5)=(10binomial5)(0.44)^5(1-0.44)^{10-5}=\frac{10!}{5!5!}(0.44)^5(0.56)^5 \\ =252(0.44)^5(0.56)^5 \\ \approx0.228878... \end{gathered}[/tex]