Determining [tex]P \quad 5 \leq X \leq 10[/tex] is 0.6822.
As per the details share in the above question are as follow,
Let us consider,
[tex]\lambda=4[/tex] days per 30 working day.
[tex]$X=$[/tex] no.of times that executive will be late to the office in next 60 working days.
Here each day is independent
Rate of arriving late at office is constant
Late day of occurrence is Random process.
[tex]& P(x)=\frac{e^{-\lambda t}(\lambda t)^x}{x !} \\& P(x)=\frac{e^{-\frac{4}{30} \times 60}\left(\frac{4}{30} \times 60\right)^x}{x !}=\frac{e^{-8}(8)^x}{x !}[/tex]
Now consider the probability,
[tex]& =p(5 \leq x \leq 10) \\& =\frac{e^{-8}(8)^5}{5 !}+\frac{e^{-8}(8)^6}{6 !}+\cdots+\frac{e^{-8}(8)^{10}}{10 !} \\& =0.0916+0.1222+0.1396+0.1396+0.1241+0.0993 \\& =0.6822[/tex]
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Note the correct question is
On experience, it is found that an executive is late for office on four days out of 30 working days. Let X denote the number of times that the executive will be late to the office in the next 60 working days. Determine [tex]P \quad 5 \leq X \leq 10[/tex].
