Respuesta :
Answer:
The probability that 10 or more of them used their phones for guidance on purchasing decisions = [tex]P(X\geq10 )[/tex] = .278
Step-by-step explanation:
Given -
A study conducted by the Pew Research Center reported that 58% of cell phone owners used their phones inside a store for guidance on purchasing decisions .
Then the probability of success is (p) = 58[tex]\%[/tex] = .58
the probability of failure is (q) = 1 - p = .42
sample size n = 15
Let X be the no of owners used their phones for guidance on purchasing decisions
Using the formula
[tex]P(X = r )= \binom{n}{r}(p)^{r}(q)^{n - r}[/tex]
The probability that 10 or more of them used their phones for guidance on purchasing decisions = [tex]P(X\geq10 )[/tex]
= P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)
= [tex]\binom{15}{10}(.58)^{10}(.42)^{15 - 10} + \binom{15}{11}(.58)^{11}(.42)^{15 - 11} + \binom{15}{12}(.58)^{12}(.42)^{15 - 12} + \binom{15}{13}(.58)^{13}(.42)^{15 - 13} + \binom{15}{14}(.58)^{14}(.42)^{15 - 14} + \binom{15}{15}(.58)^{15}(.42)^{15 - 15}[/tex]= [tex]\binom{15}{10}(.58)^{10}(.42)^{5} + \binom{15}{11}(.58)^{11}(.42)^{4} + \binom{15}{12}(.58)^{12}(.42)^{3} + \binom{15}{13}(.58)^{13}(.42)^{2} + \binom{15}{14}(.58)^{14}(.42)^{1} + \binom{15}{15}(.58)^{15}(.42)^{0}[/tex]
= [tex]\frac{15!}{(10!)(5!)}(.58)^{10}(.42)^{5} + \frac{15!}{(10!)(5!)}(.58)^{11}(.42)^{4} + \frac{15!}{(10!)(5!)}(.58)^{12}(.42)^{3} + \frac{15!}{(10!)(5!)}(.58)^{13}(.42)^{2} + \frac{15!}{(10!)(5!)}.58)^{14}(.42)^{1} + \frac{15!}{(10!)(5!)}(.58)^{15}(.42)^{0}[/tex]=
[tex]2002\times.0043\times.013 + 1365\times.0024\times.031 + 455\times.00144\times.074 + 105\times.00084\times.17 + 15\times.00048\times.42 + 1\times.00028\times1[/tex]
= .1119 + .1015 + .048 + .014 + .0030 + .00028
= .278
