Question

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. A sample of 15 cell phone owners is studied. What is the probability that 10 or more of them used their phones for guidance on purchasing decisions? Round your answer to 2 decimal places

Answer 1

The probability that 10 or more of them used their phones for guidance on purchasing decisions =
`P(X\geq10 )` = .278

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
`\%` = .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

`P(X = r )= \binom{n}{r}(p)^(r)(q)^(n - r)`

The probability that 10 or more of them used their phones for guidance on purchasing decisions =
`P(X\geq10 )`

= P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

=
`\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)`=
`\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)`

=
`(15!)/((10!)(5!))(.58)^(10)(.42)^(5) + (15!)/((10!)(5!))(.58)^(11)(.42)^(4) + (15!)/((10!)(5!))(.58)^(12)(.42)^(3) + (15!)/((10!)(5!))(.58)^(13)(.42)^(2) + (15!)/((10!)(5!)).58)^(14)(.42)^(1) + (15!)/((10!)(5!))(.58)^(15)(.42)^(0)`=

`2002*.0043*.013 + 1365*.0024*.031 + 455*.00144*.074 + 105*.00084*.17 + 15*.00048*.42 + 1*.00028*1`

= .1119 + .1015 + .048 + .014 + .0030 + .00028

= .278